Calculus I Review Solutions 2003

so that 2y = π2 , since sin(π/2) = 1. Now, y = now we need a slope: Calculus I Review Solutions 1. Compare and contrast the three “Value Theorems” of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems. The intermediate and extreme value theorems only require the function to be continuous on a closed interval. The mean value theorem additionally requires f to be differentiable on the open interval. 1 = cos(2y) · 2 and the slope at y = dy dx is dy ⇒1=0 dx This means that there is no slope- the tangent line is vertical. Therefore, the equation of the tangent line is x = 1. 5. The general solution to (y − k)0 = c(y − k) is given by: y − k = Aect y = Aect + k where A could be solved for if we were given a value for y(0). (b) The Intermediate Value Theorem says that for every y−value w between f (a) and f (b), there is an x in [a, b] so that f (x) = w. We can use this to prove the existence of solutions to f (x) = 0, if f (a) and f (b) are different in sign. 6. Compute the derivative of y with respect to x: √ √ (a) y = 3 2x + 1 5 3x − 2 Use the product rule: √ √ 1 1 y 0 = (2x+1)−2/3 ·2· 5 3x − 2+ 3 2x + 1· (3x−2)−4/5 ·3 3 5 (c) The Extreme Value Theorem tells us when we can guarantee the existence of maximums and minimums, and tells us where they occur (at endpoints or a critical point). (b) y = 1 , 1+u2 where u = In this case, 2. List the three things we need to check to see if a function f is continuous at x = a. (1) f (a) exists, (2) lim f (x) exists, and (3) Parts (1) and dy dx = dy du 1 1+x2 du · dx , dy −2u = , du (1 + u2 )2 so x→a 3. Find the point on the parabola x + y 2 = 0 that is closest to the point (0, −3). In general, the distance between a point (a, b) and (x, y) is p d= (x − a)2 + (y − b)2 so du −2x = dx (1 + x2 )2 dy 4ux = dx (1 + u2 )2 (1 + x2 )2 (2) have the same value. 1 with u = 1+x 2 , which you can either state or explicitly substitute. √ √ (c) 3 y + 3 x = 4xy Implicit differentiation: but minimizing d is equivalent to minimizing D = (x − a) + (y − b) OK, so 1 = cos(π/2) · 2 (a) We use the Mean Value Theorem for problems relating the derivative of an interior point to the values of f at the endpoints. For example, if f has two roots at x = a, x = b, then we know f must have a critical point between a and b. A common use is to substitute f 0 (c)(b − a) in place of f (b) − f (a). 2 π 4 π . 4 1 −2/3 dy 1 dy y + x−2/3 = 4y + 4x 3 dx 3 dx 2 Bring all the which is MUCH easier to differentiate. In this case,  D = (x − 0)2 + (y − (−3))2 = x2 + (y + 3)2 (d) Now we set the derivative equal to zero and solve, 4y 3 + 2y + 6 = 0 terms together: 1 −2/3 y − 4x 3 The equation of the parabola is used to get D in terms of one variable. Since x + y 2 = 0, x = −y 2 , so D = (−y 2 )2 + (y + 3)2 = y 4 + (y + 3)2 = y 4 + y 2 + 6y + 9 dy dx  dy 1 = 4y − x−2/3 ⇒ dx 3 4y − 13 x−2/3 dy = 1 dx y −2/3 − 4x 3 √ √ x+y = 3x−y Another implicit differentation: 1 1 (x + y)−1/2 (1 + y 0 ) = (x − y)−2/3 (1 − y 0 ) 2 3 (NOTE: On the exam, I wouldn’t give you a cubic equation to solve, in general, these are difficult). HINT: y = −1 is one solution. With this, we can do long division to factor out (y + 1): Multiply out so that we can isolate y 0 1 1 (x + y)−1/2 + y 0 · (x + y)−1/2 = 2 2 1 1 (x − y)−2/3 − y 0 · (x − y)−2/3 3 3 Now isolate y 0 4y 3 + 2y + 6 = 2(y + 1)(2y 2 − 2y + 3) From which we get that y = −1 is the only critical point (if you use the quadratic formula on 2y 2 − 2y + 3, you’ll get no real solutions). Now, x = −(−1)2 = −1. Is it a minimum? Yes, which you can get from looking at sign changes of the derivative. y0  1 1 (x + y)−1/2 + (x − y)−2/3 2 3  = 1 1 (x − y)−2/3 − (x + y)−1/2 3 2 Final answer: 4. Write the equation of the line tangent to x = sin(2y) at x = 1. We need a point and a slope. If x = 1, then y0 = 1 = sin(2y) 1 1 (x 3 1 (x 2 − y)−2/3 − 12 (x + y)−1/2 + y)−1/2 + 13 (x − y)−2/3 (q) y = x3−1/x (e) y = sin(2 cos(3x)) Chain Rule: y 0 = cos(2 cos(3x))·(−2 sin(3x))·3 = −6 cos(2 cos(3x)) sin(3x) (f) y = (cos(x))2x Logarithmic Differentiation: e2x ln(cos(x)) , and: Write (cos(x))2x = y 0 = e2x ln(cos(x)) (cos(x))2x (2 ln(cos(x)) + 2x tan(x)) (g) y = (tan−1 (x))−1 Chain Rule: y 0 = −(tan−1 (x))−2 · 1 p 1− (i) y = log10 (x2 2 xx +2 (j) y = entiate: (cos−1 (x))2 First rewrite: −1 2 e(x +2) ln(x) , 2 +2) ln(x) 2 y 0 = xx · (2x − 1) then differ- +2   2x ln(x) + x + 2x ln(x) + x + 2 x 2 x  7. Find the local maximums and minimums: f (x) = x3 − 3x + 1 Show your answer is correct by using both the first derivative test and the second derivative test. To find local maxs and mins, first differentiate to find critical points: so at x = −1, f is concave down, so we have a local max, and at x = 1, f is concave up, so we have a local min.  8. Compute the limit, if it exists. You may use any method (except a numerical table). (k) y = ecos(x) + sin(5x ) Chain rule: x − sin(x) x3 We have a form of (a) lim x→0 y 0 = ecos(x) (− sin(x)) + cos(5x ) · 5x ln(5) (l) y = cot(3x2 + 5) Chain rule: lim x→0 y 0 = − csc2 (3x2 + 5) · (6x) = −6x csc2 (3x2 + 5) p √ (m) y = sin( x) Chain rule: √
−1/2 √ 1 1 sin( x) · cos( x) · x−1/2 y0 = 2 2 √ √ (n) x+ 3y =1 Implicit Differentiation: 1 dy 1 −1/2 x + y −2/3 =0 2 3 dx so 3y 2/3 dy = − 1/2 dx 2x (o) x tan(y) = y − 1 Product rule/Implicit Diff We still have so use L’Hospital’s rule: x − sin(x) 1 − cos(x) = lim x3 3x2 x→0 0 , 0 so do it again and again! 1 − e−2x x→0 sec(x) 1 Note that sec(0) = cos(0) = 1, so this function is continuous at x = 0 (we can substitute x = 0 in directly), and we get that the limit is 0. x−4 (c) lim x→4+ |x − 4| Rewrite the expression to get rid of the absolute value: (b) lim x−4 = |x − 4| Solve for y 0 : y0 = 0 , 0 1 − cos(x) sin(x) cos(x) 1 lim = lim = lim = 3x2 6x 6 6 x→0 x→0 x→0 tan(y)+x sec2 (y)y 0 = y 0 ⇒ tan(y) = y 0 (1−x sec2 (y)) −2 √ 4 3, t  f 00 (x) = 6x y 0 = e(x (p) y =  For the first derivative test, set up a sign chart. You should see that 3x2 − 3 = 3(x + 1)(x − 1) is positive for x < 1 and x > 1, and f 0 (x) is negative if −1 < x < 1. Therefore, at x = −1, the derivative changes sign from positive to negative, so x = −1 is the location of a local maximum. At x = 1, the derivative changes sign from negative to positive, so we have a local minimum. For the second derivative test, we compute the second derivative at the critical points: 1 − x2 1 (x2 −x) ln(10) y0 = − x) ·√ ln(3) 1 = 3−1/x 1 + x2 x f 0 (x) = 3x2 − 3 = 0 ⇒ x = ±1 1 x2 + 1 (h) y = sin−1 (cos−1 (x)) Chain Rule: y0 = y 0 = 3−1/x +x3−1/x ln(3)·  x−4 x−4 x−4 , −(x−4) if x > 4 if x < 4 = n 1 −1 if x > 4 if x < 4 Therefore, tan(y) 1 − x sec2 (y) x−4 =1 |x − 4| (Note that the overall limit does not exist, however). lim x→4+ where t = ln(x2 ). First, note that y = −2t−3/4 and t = 2 ln(x). Now, r dy dy dt = · dx dt dx (d) 2x2 − 1 x + 8x2 For this problem, we should recall that if x < 0, √ then x = − x2 , although in this particular case, the negative signs will cancel: lim x→−∞ where dy −3 −7/4 dt 2 = t , = dt 4 dx x so put it all together (and substitute back for x): r dy −3 2 −6 = (2 ln(x))−7/4 · = dx 4 x 4x(2 ln(x))7/4 lim x→−∞ 2 2x2 − 1 · x + 8x2 −1 √ x2 −1 √ x2 s = lim x→−∞ 2− 1 x 1 x2 +8 r = 2 1 = 8 2 (e) p x2 + x + 1 − lim p x2 − x so we focus on the exponent: x→∞ Multiply by the conjugate (or rationalize): √ √ p p x2 + x + 1 + x2 − x lim x2 + x + 1− x2 − x· √ = √ x→∞ x2 + x + 1 + x2 − x 2x + 1 lim √ x→∞ lim x→∞ 2x + 1 x2 + x + 1 + √ 1 lim x 1−x = e−1 x→1 1+ 1 x 1 x2 + x2 − x p 1 x 1− + 9. Determine all vertical/horizontal asymptotes and critical 1 x √1 x2 · 1 x 2+ p points of f (x) = = =1 (1 + h)−2 − 1 h h→0 For practice, we’ll try it without using L’Hospital’s rule: (1 + h)−2 − 1 lim = lim h h→0 h→0 h→0 1 (1+h)2 lim x→±∞ = 1 − (1 + h)2 −2h − h2 = lim = −2 2 h(1 + h) h→0 h(1 + h)2 f (x) = EXTRA: This limit was the derivative of some function at some value of x. Name the function and the x value1 . First rewrite the function so that it’s in an acceptable form for L’Hospital’s rule: Note that rule: 3x2 2 2xex lim x→∞ = 3x 2ex2 x→2+ x→∞ 3 4xex2 =0 For the second part, we know that f must be continuous to be differentiable, so that leaves us with b = 4 − 2m. Also, the derivatives need to match up at x = 2. On the right side of x = 2, f 0 (x) = 2x and on the left side of x = 2, f 0 (x) = m. Therefore, 4 = m and b = 4 − 2 · 4 = −4. To be differentiable at x = 2, we require m = 4 and b = −4. x1000 − 1 (h) lim x−1 x→1 Using L’Hospital’s rule: lim x→1 x1000 − 1 1000x999 = lim = 1000 x−1 1 x→1 x tan−1 (4x) Recall that tan−1 (0) = 0, since tan(0) = 0, so this is in a form for L’Hospital’s rule: 11. Find the local and global extreme values of f (x) = x on the interval [−2, 0]. x2 +x+1 (i) lim x→0 lim x→0 x = lim tan−1 (4x) x→0 1 1 1+(4x)2 ·4 = lim x→0 x→2− For the limit to exist, we must have 4 = 2m + b. This will also automatically make item 3 true. There are an infinite number of possible solutions. Given any m, b = 4 − 2m. , and again use L’Hospital’s = lim x≤2 x>2 x→2+ x→2− 2 x→∞ 2xex 3x 2 2ex if if lim f (x) = lim mx + b = 2m + b 3x2 = lim x2 mx + b • To compute the limit, we have to do them separately: lim f (x) = lim x2 = 4 lim x e x→∞ 2 x→∞ ex n • f (2) = 2m + b, so f (2) exists. 3 −x2 x3 =2 First we see that if x < 2, f (x) = x2 which is continuous, and if x < 2, f (x) = mx + b, which is also continuous for any value of m and b. The only problem point is x = 2, so we check the three conditions from the definition of continuity: (1 + h)−2 − 1 −2(1 + h)−3 lim = lim = −2 h 1 h→0 h→0 lim 2 x2 10. Find values of m and b so that (1) f is continuous, and (2) f is differentiable. With L’Hospital: (g) 2x2 2 = lim x2 − x − 2 x→±∞ 1 − 1 − x so y = 2 is the vertical asymptote (for both +∞ and −∞). −1 h 2x2 x2 −x−2 The vertical asymptotes: x2 −x−2 = 0 ⇒ (x−2)(x+1) = 0, so x = −1, x = 2 are the equations of the vertical asymptotes (note that the numerator is not zero at these values). The horizontal asymptotes: (f) lim lim x→1 ln(x) = lim x = −1 1−x x→1 −1 so the overall limit: · √ x→∞ x2 + x + 1 + x2 − x Now divide numerator and denominator: lim √ 1 lim 1 + 16x2 1 = 4 4 We see that x2 + x + 1 = 0 has no solution, so f (x) is continuous on [−2, 0]. Therefore, the extreme value theorem is valid. Next, find the critical points: f 0 (x) = (x2 + x + 1) − x(2x + 1) −x2 + 1 = 2 2 2 (x + x + 1) (x + x + 1)2 1 (j) lim x 1−x so the critical points are x = ±1 of which we are only concerned with x = −1. Now build a chart of values: x→1 In this case, recall that x 1 1−x 1 1 lim x 1−x = lim e 1−x x→1 1 The =e ·ln(x) 1 ·ln(x) 1−x , so: x f (x) ln(x) lim 1 −x x→1 =e 0 0 −1 −1 −2 −2/3 The minimum occurs at x = −1 and the maximum occurs at x = 0. The minimum value is −1 and the maximum value is 0. x→1 function is f (x) = x−2 at x = 1 3 For the local min/max, use a sign chart for f 0 (x) in the interval [−2, 0] (or you can use the second derivative test). We see that the denominator of f 0 (x) is always positive, and the numerator is 1 − x2 , which changes from negative to positive at x = −1, so that f is decreasing then increasing and there is a local minimum at x = −1. (d) d (10x ) dx = x10x−1 False. We cannot use the Power Rule, since there is a variable in the exponent. The correct derivative is found using the rule for ax : d (10x ) = 10x ln(10) dx 12. Suppose f is differentiable so that: (e) If f 0 (x) exists and is nonzero for all x, then f (1) 6= f (0). If g(x) = f (x3 + f (x2 )), evaluate g 0 (1). True. If f 0 (x) exists for all x, then f is differUse the chain rule to get that: entiable everywhere (and is also continuous every

where). Thus, the Mean Value Theorem applies. If g 0 (x) = f 0 x3 + f (x2 ) · 3x2 + f 0 (x2 ) · 2x f (1) = f (0), that would imply the existence of a c Be careful with the parentheses!: in the interval (0, 1) so that g 0 (1) = f 0 (1+f (1))·(3+2f 0 (1)) = f 0 (1+1)(3+2) = 5f 0 (2) = 5·2 = 10 f (1) − f (0) =0 f 0 (c) = 1−0 13. Let x2 y + a2 xy + λy 2 = 0 f (1) = 1, f (2) = 2, f 0 (1) = 1 f 0 (2) = 2 but we’re told that f 0 (x) 6= 0. (a) Let a and λ be constants, and let y be a function of dy x. Calculate dx : (f) If y = ax + b, then dy dy dy 2xy + x2 + a2 y + a2 x + 2λy =0 dx dx dx dy dy −(2xy + a2 y) (x2 +a2 x+2λy) = −(2xy+a2 y) ⇒ = 2 dx dx x + a2 x + 2λy (b) Let x and y be constants, and let a be a function of da λ. Calculate dλ : 2axy da da −y 2 + y2 = 0 ⇒ = dλ dλ 2axy EXTRA2 : What is dλ ? da 14. Show that x4 + 4x + c = 0 has at most one solution in the interval [−1, 1]. We don’t need the Intermediate Value Theorem here, only the Mean Value Theorem. The derivative is 4x3 +4, so the only critical point is x = −1, which is also an endpoint. This implies: (1) If x4 +4x+c = 0 had two solutions (which is possible), then one of them must be outside the interval, since the two solutions must be on either side of x = −1. Therefore, there could be one solution inside the interval. (2) There cannot be any other solution to x4 + 4x + c = 0 inside the interval, because then there would have to be another critical point in [−1, 1]. Therefore, we conclude that there is at most one solution inside the interval (there might be no solutions). (g) If 2x + 1 ≤ f (x) ≤ x2 + 2 for all x, then limx→1 f (x) = 3. True. This is the Squeeze Theorem. If f (x) is trapped between 2x + 1 and x2 + 2 for all x, and since the limit as x → 1 of 2x + 1 is 3, and the limit as x → 1 of x2 + 2 = 3, then that forces the limit as x → 1 of f (x) to also be 3. True. The statement that f 0 (r) exists says that f is differentiable at r. The statement that limx→r f (x) = f (r) is asking if f is continuous at r. We know that all differentiable functions are continuous, so the statement is True. (i) If f and g are differentiable, then: d (f (g(x)) = f 0 (x)g 0 (x) dx False. The chain rule states that f 0 (g(x))g 0 (x) d (f (g(x)) dx = (j) If f (x) = x2 , then the equation of the tangent line at x = 3 is: y − 9 = 2x(x − 3) False. 2x is a formula for the slope, not the slope itself. The equation of the tangent line is: y − 9 = 6(x − 3). (k) limπ θ→ 3 cos(θ) − θ− π 3 1 2 = − sin π 3   True by l’Hospital’s rule. You could have also said that the expression is in the form: (c) If f is differentiable, then d p f 0 (x) f (x) = p dx 2 f (x) lim x→a True, since f (x) − f (a) x−a with f (x) = cos(x) and a = π3 . This is another way of defining the derivative of cos(x) at x = π3 . d p 1 f 0 (x) f (x) = (f (x))−1/2 · f 0 (x) = p dx 2 2 f (x) = dy =x+0=x da lim f (x) = f (r) (b) If f (x) is decreasing and g(x) is decreasing, then f (x)g(x) is decreasing. False. The derivative is found via the product rule, f 0 g + f g 0 . We know that f 0 (x) < 0 and g 0 (x) < 0, but we do not know the signs of f (x) and g(x), so we cannot say that the derivative of f g is negative. dλ da dy True. If we’re computing da , then we’re treating x and b as constants. Differentiating, we get x→r (a) If f has an absolute minimum at c, then f 0 (c) = 0. False. For example, f (x) = |x| has an absolute minimum at x = 0, but f 0 (x) is not defined at x = 0. answer is =x (h) If f 0 (r) exists, then 15. True or False, and give a short explanation. 2 The dy da (l) There is no solution to ex = 0 True. If there were a solution, it would be x = ln(0), but ln(0) is not defined. 2axy −y 2 4 2π 3 (m) sin−1 sin

2π 3 = Another way to say this: What are the tangent lines through y = 8 − x2 that also go through (4, 0)? The unknown value here is the x−coordinate, so let x = a. Then the slope is −2a, and the corresponding point on the curve is (a, 8 − a2 ). The general form of the equation of the tangent line is then given by: False, with the usual restrictions on the sine function. That is, if − π2 ≤ θ ≤ π2 , then it is true that sin−1 (sin(θ)) = θ. In this case,  sin−1 sin  2π 3  = π 3 y − 8 + a2 = −2a(x − a) 5log5 (2x) (n) = 2x, for x > 0. True, since 5x and log5 (x) are inverses of each other. We needed x > 0 so that log5 (2x) is defined. where x, y are points on the tangent line. We want the tangent line to go through (4, 0), so we put this point in and solve for a: (o) d ln(|x|) = dx True: −8 + a2 = −2a(4 − a) = −8a + 2a2 ⇒ √ 8 ± 32 ⇒ a= 2 1 , x for all x 6= 0. ln |x| = n ln(x), ln(−x), if x > 0 if x < 0 so so we take the leftmost value, a = d ln |x| = dx (p)  1 , x 1 −x · (−1) = 1 , x = x10x−1 Oops- a duplication. See part (c) (q) If x > 0, then (ln(x))6 = 6 ln(x) False. The rule says: log(ab ) = b log(a), but here the 6 is outside the logarithm. 1 1 = lim √ = 6 x→7 ( x + 2 + 3) √ which is the derivative of x + 2 at x = 7. 16. Find the domain of ln(x − x2 ): Use a sign chart to determine where x−x2 = x(1−x) > 0: − + x<0 + + 0 0). We can use the second derivative test to check these out: At x =p 0, f 00 (0) = p −4, so x = 0 is always a local max. At x = ± 1/c, f 00 (± 1/c) = 12 − 4 = 8. So, if c > 0, there + − x>1 so overall, x − x2 > 0 if 0 < x < 1. 17. Find the value of c guaranteed by the Mean Value Theox rem, if f (x) = x+2 on the interval [1, 4]. To set things up, we see that f is continuous on [1, 4] and differentiable on (1, 4), since the only “bad point” is 2 1 x = −2. We should get that f 0 (x) = (x+2) 2 , f (1) = 3 p are local mins at x = ± 1/c. √ 23. If f (x) = 1 − 2x, determine f 0 (x) by using the definition of the derivative. p √ 1 − 2(x + h) − 1 − 2x 0 = f (x) = lim h h→0 p p √ √ 1 − 2(x + h) + 1 − 2x 1 − 2(x + h) − 1 − 2x ·p lim = √ h h→0 1 − 2(x + h) + 1 − 2x and f (4) = 23 . Therefore, the Mean Value Theorem says that c should satisfy: 2 1 −3 1 2 = = 3 4−1 9 (c + 2)2 or √ (c + 2)2 = 18 ⇒ c = −2 ± 18 √ of which only −2 + 18 ≈ 2.243 is inside our interval. 1 − 2x − 2h − 1 + 2x = √ h( 1 − 2(x + h) + 1 − 2x) lim 18. Given that the graph of f passes through the point (1, 6) and the slope of the tangent line at (x, f (x)) is 2x + 1, find f (2). Since f 0 (x) = 2x + 1, f (x) = x2 + x + C is the general antiderivative. Given that (1, 6) goes through f , 12 + 1 + C = 6 ⇒ C = 4. Therefore, f (x) = x2 + x + 4. Now, f (2) = 4 + 2 + 4 = 10. p h→0 lim h→0 −2 p 1 − 2(x + h) + √ = √ 1 − 2x −1 1 − 2x 24. A point of inflection for a function f is the x value for which f 00 (x) changes sign (either from positive to negative or vice versa). 19. To simplify, we make θ = tan−1 (x) so that tan(θ) = x1 . This means we are looking at a triangle whose length of the side√opposite is x, and adjacent is 1. The hypotenuse is then 1 + x2 . Now take the cos(θ) to get: cos(tan−1 (x)) = √ √ 8− 32 . 2 21. Compute the limit, without using L’Hospital’s Rule. √ x+2−3 lim x−7 x→7 Rationalize to get: √ √ x+2−3 x+2+3 x+2−9 ·√ lim = lim √ x−7 x→7 x→7 (x − 7)( x + 2 + 3) x+2+3 if x > 0 if x < 0 d 10x dx x 1−x 0 = a2 − 8a + 8 If f 00 is continuous, and f 00 (a) < 0 and f 00 (b) > 0, there is a c so that f 00 (c) = 0 is an inflection point. 1 ...which is the Intermediate Value Theorem. Find constants a and b so that (1, 6) is an inflection point for y = x3 + ax2 + bx + 1. Differentiate twice to get: 1 + x2 20. A fly is crawling from left to right along the curve y = 8 − x2 , and a spider is sitting at (4, 0). At what point along the curve does the spider first see the fly? y 00 = 6x + 2a 5 30. If f (x) = (2 − 3x)−1/2 , find f (0), f 0 (0), f 00 (0). Differentiate: At x = 1, we want an inflection point, so 6 + 2a should be a point where y 00 changes sign: 6 + 2a = 0 ⇒ a = −3. We see that if a < −3, then y 00 < 0, and if a > −3, y 00 > 0. Putting this back into the function, we have: f 0 (x) = y = x3 − 3x2 + bx + 1 −9 27 (2 − 3x)−5/2 (−3) = (2 − 3x)−5/2 4 4 √ √ Now, (note that 23/2 = 2 2 and 25/2 = 4 2): f 00 (x) = and we want the curve to go through the point (1, 6): 6=1−3+b+1 1 f (0) = √ , 2 so b = 7. 25. Suppose that F (x) = f (g(x)) and g(3) = 6, g 0 (3) = 4, f (3) = 2 and f 0 (6) = 7. Find F 0 (3). By the Chain Rule: so F 0 (3) = f 0 (6) · 4 = 7 · 4 = 28 26. Find the dimensions of the rectangle of largest area that has its base on the x−axis and the other two vertices on the parabola y = 8 − x2 . Try drawing a picture first: The parabola opens down, goes √ through the y−intercept at 8, and has x−intercepts of ± 8. Now, let x be as usual, so that the full length of the base of the rectangle is 2x. Then the height is y, or 8 − x2 . Therefore, the area of the rectangle is: 0.5 · and solve for 1 1 y − 1 = − x, or y = − + 1 2 2 √ 33. Find f (t), if f 00 (t) = t + t, and f (1) = 1, f 0 (1) = 2. f 0 (t) = 1 2 2 t + t3/2 + C 2 3 so f 0 (1) = 2 means: 1 2 5 + + C = 2, so C = 2 3 6 Now, f 0 (t) = 2 f (t) = 12t − 12t = 0 ⇒ t = 0, t = 1 f (t) = The velocity at t = 0 is 0 and the velocity at t = 1 is 4 − 6 = −2. √ 29. If y = 5t − 1, compute y 000 . Nothing tricky here- Just differentiate, and differentiate, and differentiate! y 00 = 1 (1 − x)−1/2 (−1) 2 so f 0 (0) = − 12 , and the point is (0, 1). 28. If position is given by: f (t) = t4 − 2t3 + 2, find the times when the acceleration is zero. Then compute the velocity at these times. Take the second derivative, and set it equal to zero: y0 = dz , 78. dt f 0 (x) = √ 27. Let G(x) = h( x). Then where is G differentiable? Find 0 G (x). √ First compute G0 (x) = h0 ( x) 12 x−1/2 . From this we see that as long as h is differentiable and x > 0, then G will be differentiable. f (x) = 4t − 6t , dz = 0.3 · 50 + 0.4 · 60 dt √ 32. Find the linearization of f (x) = 1 − x at x = 0. To linearize, we find the equation of the tangent line. 8 1 2x = √ , y = 5 3 6 00 27 √ 16 2 Translating the question, we get that we want to find dz dt when A = 0.3, B = 0.4, (so z = 0.5), A0 (t) = 50, B 0 (t) = 60. Then: dz dA dB = 2A + 2B 2z dt dt dt The two’s divide out and put in the numbers: dA = 16 − 6x2 dx so the critical points are x = ± √4 , of which only x = √4 6 6 is in our interval. So the dimensions of the rectangle are: 2 f 00 (0) = z 2 = A2 + B 2 A = 2xy = 2x(8 − x2 ) = 16x − 2x3 √ and 0 ≤ x ≤ 8. We see that the area will be zero at the endpoints, so we expect a maximum at the critical point inside the interval: 3 3 1 3 · = √ , 2 23/2 4 2 f 0 (0) = 31. Car A is traveling west at 50 mi/h, and car B is traveling north at 60 mi/h. Both are headed for the intersection between the two roads. At what rate are the cars approaching each other when car A is 0.3 mi and car B is 0.4 mi from the intersection? Let A(t), B(t) be the positions of cars A and B at time t. Let the distance between them be z(t), so that the Pythagorean Theorem gives: F 0 (3) = f 0 (g(3))g 0 (3) 0 3 −1 (2 − 3x)−3/2 (−3) = (2 − 3x)−3/2 2 2 1 2 t 2 + 23 t3/2 + 56 , and 1 1 3 2 2 5/2 5 1 4 5 · t + · t + t+C = t3 + t5/2 + t+C 2 3 3 5 6 6 15 6 Now, f (1) = 1 means: 1 4 5 5 + 8 + 25 19 −4 + + +C = 1 ⇒ +C = 1 ⇒ C = 1− = 6 15 6 30 15 15 34. Find f 0 (x) directly from the definition of the derivative (using limits and without L’Hospital’s rule): First, recall that: 1 5 (5t − 1)−1/2 · 5 = (5t − 1)−1/2 2 2 5 −1 −25 · (5t − 1)−3/2 · 5 = (5t − 1)−3/2 2 2 4 375 y 000 = (5t − 1)−5/2 8 f 0 (x) = lim h→0 f (x + h) − f (x) h We’ll use this for these exercises: 6 √ 40. (a) f (x) = 3 − 5x √ √ √ √ 3 − 5x − 5h − 3 − 5x 3 − 5x − 5h + 3 − 5x lim ·√ √ h h→0 3 − 5x − 5h + 3 − 5x 3 − 5x − 5h − 3 + 5x
√ √ h 3 − 5x − 5h + 3 − 5x lim h→0 Show that 5 is a critical number of g(x) = 2 + (x − 5)3 , but that g does not have a local extremum there. g 0 (x) = 3(x − 5)2 , so g 0 (5) = 0. By looking at the sign of g 0 (x) (First derivative test), we see that g 0 (x) is always non-negative, so g does not have a local min or max at x = 5. 41. Find the general antiderivative: (SORRY, THIS SHOULD NOT HAVE BEEN INCLUDED HERE- IGNORE THIS PROBLEM FOR NOW) −5 2 3 − 5x √ (b) f (x) = x2 lim h→0 (a) f (x) = 4 − x2 + 3ex F (x) = 4x − 13 x3 + 3ex + C (x + h)2 − x2 x2 + 2xh + h2 − x2 = lim = h h h→0 (b) f (x) = (c) f (x) = h(2x + h) lim = 2x h h→0 3 + x2 x2 1+x √ x First rewrite f (x) = x−1/2 + x1/2 , and 2 3/2 x +C 3 42. Find the slope of the tangent line to the following at the √ point (3,4): x2 + yx + y 2 = 31 F (x) = 2x1/2 + (c) f (x) = x−1 1 x+h lim 1 x − = lim h h→0 x−(x+h) x(x+h) h h→0 = 2x + −h −1 −1 lim = lim = 2 x h→0 hx(x + h) h→0 x(x + h) f 0 (f (f (f (x)))) · f 0 (f (f (x))) · f 0 (f (x)) · f 0 (x) which simplifies (since f (0) = 0) to: f 0 (0) · f 0 (0) · f 0 (0) · f 0 (0) = 24 = 16 36. Differentiate: f (x) = n √ x √ − x f (x) = if x ≥ 0 if x < 0 f 0 (x) = 37. f (x) = | ln(x)|. Find f 0 (x). We can rewrite f (Recall that ln(x) < 0 if 0 < x < 1)  0 √ x) . f (x) = e Find f 0 (x). g( √ x) + xe g( 2x − 1 −2x + 1 Line: y − 1 = √ x) √ 1 · g ( x) · x−1/2 2 0 f (2) − f (0) 5 = 2−0 2 1 (x 2 − 0), or y = 1 x 2 +1 46. (THIS PROBLEM SHOULD HAVE BEEN DELETED√ SORRY!) Find dy if y = 1 − x and evaluate dy if x = 0 and dx = 0.02. Compare your answer to ∆y 39. Find a formula for dy/dx: x2 + xy + y 3 = 0. 2x+y+xy 0 +3y 2 y 0 = 0 ⇒ (x+3y 2 )y 0 = −(2x+y) ⇒ y 0 = if x < 0, or x > 1 if 0 < x < 1 Since 52 > 2, there can exist no function like that (that is continuous). √ 45. Linearize f (x) = 1 + x at x = 0. Point: x = 0, y = 1 Slope: f 0 (0) = 12 if x > 1 if 0 < x < 1 Note that the pieces don’t match at x = 1; we remove that point from the domain. 38. f (x) = xeg( n f 0 (x) = 1 , x − x1 , if x ≤ 0, or x ≥ 1 if 0 < x < 1 44. Does there exist a function f so that f (0) = −1, f (2) = 4, and f 0 (x) ≤ 2 for all x? if x ≥ 1 if 0 < x < 1 and differentiate piecewise: f 0 (x) = x2 − x −x2 + x At x = 0, from the left, f 0 (x) → 1 and from the right, f 0 (x) → −1, so f 0 (x) does not exist at x = 0. At x = 1, from the left, f 0 (x) → −1, and from the right, f 0 (x) → 1, so f 0 (x) does not exist at x = 1. Finally, f 0 (x) = 0 if 2x − 1 = 0 ⇒ x = 1/2, but 1/2 is not in that domain. The other part is where −2x + 1 = 0, which again is 1/2, and this time it is in 0 < x < 1. The critical points are: x = 1/2, 0, 1. If x < 0, f 0 (x) = 2√1−x , which also goes to infinity as x approaches 0 (from the left). ln(x), − ln(x), n Now compute the derivative: Is f differentiable at x = 0? Explain. √ √ Correction: − x = − −x Is f differentiable at x = 0? Explain. f will not be differentiable at x = 0. Note that, if x > 0, 1 then f 0 (x) = 2√ , so f 0 (x) → ∞ as x → 0+ x n 1 −1/2 0 √ y y x + y + 2yy 0 = 0 2 At x = 3, y = 4: −32 3 6 + y 0 + 2 + 8y 0 = 0 ⇒ y 0 = 4 35 −32 y−4= (x − 3) 35 43. Find the critical values: f (x) = |x2 − x| One way to approach this problem is to look at it piecewise. Use a table to find where f (x) = x(x − 1) is positive or negative: 35. If f (0) = 0, and f 0 (0) = 2, find the derivative of f (f (f (f (x)))) at x = 0. First, note that the derivative is (Chain Rule): f (x) = + 1 F (x) = −3x−1 + 2 ln |x| + x + C −(2x + y) x + 3y 2 7 −1 1 dy = √ dx, ⇒ dy = √ · 0.02 = 0.01 2 1−x 2 1−0 √ √ ∆y = 1 − 0.02 − 1 = 0.01005... 47. Fill in the question marks: If f 00 is positive on an interval, then f 0 is INCREASING and f is CONCAVE UP. 52. Define the functions as piecwise defined functions: |3x + 2| = 3x + 2 48. If f (x) = x − cos(x), x is in [0, 2π], then find the value(s) of x for which x−2 x+1 x+2 3π 2 0 −1 3π 2 3 2 π (b) f (x) is increasing most rapidly. Another way to say this: Where’s the maximum of f 0 (x)? We’ve computed f 0 (x) to be: 1 + sin(x), so take its derivative: cos(x) = 0. So there are two critical points at x = π2 and x = 3π . Checking 2 these and the endpoints: 0 1 π 2 3π 2 2 0 2π 1 − − + −2 < x < −1 π . 2 x→2− if − 2 < x < −1 or x ≥ 2 if x < −2 or − 1 < x < 2 lim f (x) = lim cx + d = d x→0+ Therefore, for the limit to exist at x = 0, we must have d = −1. So far, this function is now continuous at x = 0 for any value of c. At x = 1, (i) f (1) = c · 1 + d = c · 1 − 1 = c − 1, which exists for all values of c. (ii) Check that the limit exists at x = 1, lim f (x) = lim cx − 1 = c − 1 x→1− x→1− lim f (x) = lim x→1+ √ x→1+ x+3=2 Therefore, for the limit to exist at x = 1, c − 1 = 2 or c = 3. At this value of c, the third condition of continuity is also satisfied. Final answer: c = 3, d = −1. 54. Where is f continuous? Use a sign chart, since the expression under the radical sign must be nonnegative (positive or zero): x−1 x+2 x−2 50. Sketch the graph of a function that satisfies all of the given conditions: x→2+ + + + x>2 x→0− x→0+ 49. Show there is exactly one root to: ln(x) = 3 − x First, to use the Intermediate Value Theorem, we’ll get a function that we can set to zero: Let f (x) = ln(x) − 3 + x. Then a root to ln(x) = 3 − x is where f (x) = 0. First, by plugging in numbers, we see that f (2) < 0 and f (3) > 0. There is at least one solution in the interval [2, 3] by the Intermediate Value Theorem. Now, is there more than one solution? f 0 (x) = x1 + 1 which is always positive for positive x. This means that, for x > 0, f (x) is always increasing. Therefore, if it crosses the x−axis (and it does), then f can never cross again. f (4) = 2 lim f (x) = 3 if x > − 23 if x < − 23 lim f (x) = lim 2x2 − 1 = −1 (c) The slopes of the lines tangent to the graph of f are increasing most rapidly. Another way to say this: Where is f 0 (x) increasing most rapidly? At the maximum of f 00 (x). The maximum of cos(x) in the interval [0, 2π] occurs at x = 0 and x = 2π. f (1) = 5 lim f (x) = ∞, 1 −1 − + + −1 < x < 2 x−2 (x+1)(x+2) x−2 − (x+1)(x+2)  x→0− so f is increasing most rapidly at x =  = 53. Find all values of c and d so that f is continuous at all real numbers. First, we should see that the only two “problem” points are at x = 0 and at x = 1, which are where the functions come together. Other than those points, f is continuous. At the problem points, we need to check the three conditions for continuity. At x = 0, (i) f (0) = c · 0 + d, so f (0) = d which exists for all c and d. (ii) Check that the limit exists: so f is greatest at x = 2π, least at 0. x f 0 (x) − − − x < −2 x−2 = (x + 1)(x + 2) 2π 2π − 1 ≈ 5.2 ≈ 4.7 if 3x + 2 > 0 if 3x + 2 < 0 The function is positive in the second and 4th intervals, and negative in the first and third. Now, is the only critical point in [0, 2π]. Now the table: x f (x) 3x+2 3x+2 − 3x+2 3x+2 To do the next problem, we need to know where the function is positive and where it is negative. Use a sign chart: (a) f (x) is greatest and least. Here we are looking for the maximum and minimumuse a table with endpoints and critical points. To find the critical points, f 0 (x) = 1 + sin(x) = 0 ⇒ sin(x) = −1 ⇒ x =  f 0 (1) = f 0 (4) = 0 f (2) = 4 − − − x < −2 Therefore, f (x) = (We’ll do this one in class) − + − −2 < x < 1 q x−1 x2 −4 + + − 12 is continuous on its domain, which is: dt 51. If s2 t + t3 = 1, find ds and ds . dt First, treat t as the function, s as the variable: −2 < x ≤ 1 or x > 2 55. Recall our setup- we want to let Q(t) be the kilograms of salt at time t (NOT the concentration- that would require a different equation). The differential equation is then: dt dt dt −2st 2st + s + 3t2 =0⇒ = 2 ds ds ds s + 3t2 2 dQ = Rate coming in − Rate going out dt For s as the function, t as the variable: ds −(s2 + 3t2 ) = dt 2st The rate coming in is 25 which you can either state directly or show. 8 liters kg kg · 0.03 = 0.75 minute liter min 58. The snowball equation is: To compute the rate going out, there is Q kilograms of salt in 5000 liters of water, so Q/5000 is the concentration at time t. Therefore, the rate of salt leaving the tank is: 25 A = 4πr 2 where A and r are functions of t. We need to write A in terms of the diameter instead of in terms of the radius. Let h be the diameter, h = 2r and substitute: liters Q kg kg · = 0.005Q min 5000 liters min Now we have:  Q0 = 0.75−0.005Q = −0.005 Q − 0.75 0.005  A = 4π = −0.005(Q−150) h2 = πh2 4 Differentiating with respect to time, we get: which we write as: 0 (Q − 150) = −0.005(Q − 150) ⇒ Q(t) = Ae −0.005t dA dh = 2πh dt dt + 150 We solve for A by noting that Q(0) = 20, so that We are told that Q(t) = −130e−0.005t + 150 dA dt = −1, h = 10 and solve for dh , dt dh 1 =− ≈ −0.0159 dt 20π Now, after a half an hour (t = 30), we get Q(30) ≈ 38.11. 56. It might be interesting to compare the results of this problem if we assume that (i) the vertex is up, versus (ii) the vertex is down. (An inverted cone has its vertex up, so only (i) would be the solution to the problem as given): 59. For this tank mixing problem, we have: dQ Q 1 = 10 · 0 − 10 · =− Q dt 1000 100 (a) In this case, let r be the radius on the surface of the 1 so the general solution is Q(t) = Ae− 100 t . Using the water level at height h. Then the volume of water initial condition that Q(0) = 15, we get: in the cone is given by the full volume, and subtract the empty cone on the top: 1 Q(t) = 15e− 100 t π π 2 2 V = π · 2 · 4 − πr h 3 3 The mathematical model of Q says that there will never be zero salt in any finite amount of time (although in practice, Now we need to write V in terms of h, since we are the amount is negligible). trying to determine dh . From similar triangles, we dt get: 2 r 4−h = r= 4 4−h 2 Now the volume becomes 16π π 3 dV π dh − (h −8h2 +16h) = − ·(3h2 −16h+16) V = 3 12 dt 12 dt Now substitute dV dt = 2, h = 3 to find that dh dt = 24 5π (b) If the cone had its vertex at the bottom, the problem simplifies quite a bit. The volume of water is just the volume of the small cone, π V = r2 h 3 where by similar triangles we get h = 2r. Substituting like we did earlier, V = Substitute dV dt π 3 h 12 dV π dh = h2 · dt 4 dt = 2 and h = 3 to get dh dt = 8 9π 57. We did something similar to this problem in class. Form a right triangle where the hypotenuse is 25. The other two sides have length x(t) and y(t) (for specificity, let x(t) be the distance from the base of the wall to the base of the ladder, and y(t) be the height of the ladder against the wall). Then, x2 + y 2 = 625 ⇒ 2x dx dy + 2y =0 dt dt Now, we are given that dy = −1 and x = 6, so we still dt need y in order to solve for dx . From our original equation, dt if x = 6, then √ y 2 = 252 − 62 = 589 ⇒ y = 589 Now substitute these values in, and: 12 √ dx dx − 2 589 = 0 ⇒ = dt dt √ 589 6 9