Calculus I Review Solutions 2002

Calculus I Review Solutions 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). 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. Now, x = −(−1)2 = −1. Is it a minimum? Yes, which you can get from looking at sign changes of the derivative. 5. 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 (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). 1 = sin(2y) π , 2 so that 2y = since sin(π/2) = 1. Now, y = now we need a slope: 1 = cos(2y) · 2 (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. 1 00 y − 2y 0 + y = 3x2 + 2x + 1 2 x→a To start, we need expressions for y, y 0 , y 00 which we will then substitute into the equation: 3. Derive the formula for the derivative of y = sec−1 (x). First, write the expression as sec(y) = x. Now draw a hyp triangle with y one of the angles. Since sec(y) = , the adj hypotenuse is x, the side adjacent is 1 and the third side √ is x2 − 1. Now, differentiate sec(y) = x: Let’s re-write this expression by combining the x2 , x and constants: Ax2 + (B − 4A)x + (C + A − 2B) = 3x2 + 2x + 1 dy =1 dx From which we get: A B − 4A C + A − 2B Use logarithmic differentiation (its easier): 2 D = (x − 0) + (y − (−3)) = x + (y + 3) ln(y) = 2 D = (−y ) + (y + 3) = y + (y + 3) 1 1 ln(2x + 1) + ln(3x − 2) 3 5 so that The equation of the parabola is used to get D in terms of one variable. Since x + y 2 = 0, x = −y 2 , so 4 = 2 from x term = 1 from constants 7. Compute the derivative of y with respect to x: √ √ (a) y = 3 2x + 1 5 3x − 2 which is MUCH easier to differentiate. In this case, 2 = 3 from x2 term from which we get A = 3, B = 14, C = 26. D = (x − a)2 + (y − b)2 1 0 1 1 1 1 y = · ·2+ · ·3 y 3 2x + 1 5 3x − 2 2 Now, Now we set the derivative equal to zero and solve. With the hint from class, we write: 3 = 2A
1 (2A) − 2 (2Ax + B) + Ax2 + Bx + C = 3x2 + 2x + 1 2 but minimizing d is equivalent to minimizing 2 = 2Ax + B so we get: 4. 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 2 2 = Ax2 + Bx + C 0 y dy 1 = √ dx x x2 − 1 2 y y 00 dy =1 sec(y) tan(y) dx √ opp x2 −1 and sec(y) = x, tan(y) = = so 1 adj so dy ⇒1=0 dx 6. For what values of A, B, C will y = Ax2 + Bx + C satisfy the differential equation: (2) have the same value. x2 − 1 · dy dx This means that there is no slope- the tangent line is vertical. Therefore, the equation of the tangent line is x = 1. 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 p OK, so is 1 = cos(π/2) · 2 (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). x· π 4 and the slope at y = π . 4 y0 = 2 4y + 2y + 6 = 2(y + 1)(2y − 2y + 3) 1 √ 3
 √ 2x + 1 5 3x − 2 2 3 + 3(2x + 1) 5(3x − 2)  (b) y = 1 , 1+u2 1 1+x2 dy du · , du dx (h) y = sin−1 (cos−1 (x)) Chain Rule: where u = dy dx In this case, = −2u dy = , du (1 + u2 )2 so so 1 with u = 1+x 2 , which you can either state or explicitly substitute. √ √ (c) 3 y + 3 x = 4xy Implicit differentiation:  (d) 1 −2/3 y − 4x 3  Now, 10y = x2 − x, so: (x2 − x) ln(10) · y 0 = 2x − 1 terms together: and 2x − 1 (x2 − x) ln(10) Alternatively, we can compute the derivative of log10 (x): If y = log10 (x), then 10y = x, and 10y ln(10) · y 0 = 1. Now, 10y = x, so 4y − 13 x−2/3 dy 1 dy = 4y− x−2/3 ⇒ = 1 dx 3 dx y −2/3 − 4x 3 y0 = x ln(10) · y 0 = 1 ⇒ y 0 = 1 1 (x + y)−1/2 (1 + y 0 ) = (x − y)−2/3 (1 − y 0 ) 2 3 y0 = 1 1 (x + y)−1/2 + y 0 · (x + y)−1/2 = 2 2  2 ln(y) = (x2 + 2) ln(x) = Use the product rule: 1 0 1 2 y = (2x) ln(x) + (x2 + 2) = 2x ln(x) + x + y x x 1 1 (x − y)−2/3 − (x + y)−1/2 3 2 Final answer: 1 (x 3 1 (x 2 y0 = − y)−2/3 − + y)−1/2 + 1 (x 2 1 (x 3 + Final answer: y)−1/2 2 y 0 = xx +2  2x ln(x) + x + − y)−2/3 2 x  (k) y = ecos(x) + sin(5x ) Chain rule: (e) y = sin(2 cos(3x)) Chain Rule: y 0 = ecos(x) (− sin(x)) + cos(5x ) · 5x ln(5) y 0 = cos(2 cos(3x)) · (−2 sin(3x)) · 3 = (l) y = cot(3x2 + 5) Chain rule: −6 cos(2 cos(3x)) sin(3x) (f) y = (cos(x))2x Logarithmic Differentiation: y 0 = − csc2 (3x2 + 5) · (6x) = −6x csc2 (3x2 + 5) p √ (m) y = sin( x) Chain rule: ln(y) = 2x · ln(cos(x)) ⇒ 1 0 1 y = 2 ln(cos(x)) + 2x · sin(x) = y cos(x) y0 = 2 ln(cos(x)) + 2x tan(x) so that (n) y 0 = (cos(x))2x (2 ln(cos(x)) + 2x tan(x)) √
−1/2 √ 1 1 sin( x) · cos( x) · x−1/2 2 2 √ √ x+ 3y =1 Implicit Differentiation: 1 −1/2 1 dy x + y −2/3 =0 2 3 dx (g) y = (tan−1 (x))−1 Chain Rule: 0 1 · 2x − 1 (x2 − x) ln(10) (j) y = xx +2 Logarithmic Differentiation: 1 1 (x − y)−2/3 − y 0 · (x − y)−2/3 3 3 Now isolate y 0 1 1 (x + y)−1/2 + (x − y)−2/3 2 3 1 x ln(10) and now we can use the chain rule: Multiply out so that we can isolate y 0 y −1 1 − x2 10y ln(10) · y 0 = 2x − 1 √ x+y = 3x−y Another implicit differentation:  ·√ Now differentiate implicitly: √ 0 1 − (cos−1 (x))2 10y = x2 − x 1 −2/3 dy 1 dy y + x−2/3 = 4y + 4x 3 dx 3 dx dy dx p (i) y = log10 (x2 − x) First, we don’t know the derivative of log10 (x), so we can re-write the problem as: dy 4ux = dx (1 + u2 )2 (1 + x2 )2 Bring all the 1 y0 = −2x du = dx (1 + x2 )2 so −1 y = −(tan (x)) −2 1 · 2 x +1 2 dy 3y 2/3 = − 1/2 dx 2x x − sin(x) x3 We have a form of (o) x tan(y) = y − 1 Product rule/Implicit Diff (a) lim x→0 tan(y)+x sec2 (y)y 0 = y 0 ⇒ tan(y) = y 0 (1−x sec2 (y)) Solve for y 0 : lim y0 = (p) y = −2 √ 4 3, t where t = x→0 tan(y) 1 − x sec2 (y) We still have ln(x2 ). lim First, note that y = −2t−3/4 and t = 2 ln(x). Now, x→0 dy dy dt = · dx dt dx (b) lim x→0 dy −3 −7/4 dt 2 = t , = dt 4 dx x so put it all together (and substitute back for x): (c) dy −3 2 −6 = (2 ln(x))−7/4 · = dx 4 x 4x(2 ln(x))7/4 (q) y = x3−1/x y 0 = 3−1/x + x3−1/x · 1 1 = 3−1/x 1 + x2 x 1 − cos(x) x − sin(x) = lim x3 3x2 x→0 0 , 0 so do it again and again! 1 − cos(x) 1 sin(x) cos(x) = lim = lim = 3x2 6x 6 6 x→0 x→0 1 − e−2x sec(x) x−4 |x − 4| Rewrite the expression to get rid of the absolute value: lim x→4+ x−4 = |x − 4|   x−4 x−4 x−4 , −(x−4) if x > 4 if x < 4 n = 1 −1 if x > 4 if x < 4 Therefore, x2x+1 . x−4 =1 |x − 4| lim 8. Let f (x) = Without explicitly computing the inverse, what is the equation of the tangent line to f −1 (x) at x = 4? HINT: The point (1, 4) goes through the graph of f . The general rule here is the following: If (a, b) is on the graph of f , then (b, a) is on the graph of f −1 . We also said that the derivative of f −1 at b is f 01(a) , given that f 0 (a) 6= 0. So, in this problem, x→4+ (Note that the overall limit does not exist, however). r (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→−∞ f 0 (x) = 2x+1 + x2x+1 ln(2) · 1 r and f 0 (1) = 4 + 4 ln(2), so the derivative of f −1 at 4 is 1 . Now we have a slope and a point: 4+4 ln(2) y−1= so use L’Hospital’s rule: 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. where  0 , 0 lim x→−∞ 1 (x − 4) 4 + 4 ln(2) (e) 2x2 − 1 · x + 8x2 p x2 + x + 1 − lim s −1 √ x2 −1 √ x2 = 2− lim 1 x x→−∞ r 1 x2 = +8 2 1 = 8 2 p x2 − x 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 is the tangent line to f −1 at x = 4. x3 9. Find the local maximums and minimums: f (x) = − 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: 2x + 1 · √ x2 + x + 1 + x2 − x Now divide numerator and denominator: lim √ x→∞ f 0 (x) = 3x2 − 3 = 0 ⇒ x = ±1 lim √ 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: x→∞ lim x→∞ 2x + 1 x2 + x + 1 + √ 1+ 1 x + 1 x √1 x2 1 x 2+ p x2 − x · 1 x2 + p 1− 1 x = =1 (1 + h)−2 − 1 h h→0 For practice, we’ll try it without using L’Hospital’s rule: (f) lim f 00 (x) = 6x 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. lim h→0 10. Compute the limit, if it exists. You may use any method (except a numerical table). (1 + h)−2 − 1 = lim h h→0 lim h→0 3 1 (1+h)2 −1 h h2 −2h − = −2 h(1 + h)2 = lim h→0 1 − (1 + h)2 = h(1 + h)2 With L’Hospital: lim (1 + h)−2 −1 h h→0 = lim h→0 h)−3 −2(1 + 1 12. Find values of m and b so that (1) f is continuous, and (2) f is differentiable. = −2 f (x) = EXTRA: This limit was the derivative of some function at some value of x. Name the function and the x value1 . (g) 2 lim x3 e−x First rewrite the function so that it’s in an acceptable form for L’Hospital’s rule: lim x→∞ Note that rule: 3x2 2 2xex 3x 2 2ex = 3x lim x→∞ 2e x→∞ x2 x→2+ , and again use L’Hospital’s = lim x→∞ 3 4xe x2 x→2− =0 x1000 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. 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: lim x→0 x = lim tan−1 (4x) x→0 1 1 1+(4x)2 13. Find the local and global extreme values of f (x) = x on the interval [−2, 0]. x2 +x+1 1 + 16x2 1 = = lim 4 4 x→0 ·4 1 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: (j) lim x 1−x x→1 In this case, recall that x lim x 1 1−x = lim e x→1 1 1−x =e 1 ·ln(x) 1−x 1 ·ln(x) 1−x lim = ex→1 f 0 (x) = , so: x f (x) so we focus on the exponent: ln(x) = lim x = −1 1−x x→1 −1 so the overall limit: 1 lim x 1−x = e−1 x→1 11. Determine all vertical/horizontal asymptotes and critical points of f (x) = 2x2 x2 −x−2 lim If g(x) = f (x3 + f (x2 )), evaluate g 0 (1). Use the chain rule to get that:
g 0 (x) = f 0 x3 + f (x2 ) · 3x2 + f 0 (x2 ) · 2x 2 x2
=2 Be careful with the parentheses!: g 0 (1) = f 0 (1+f (1))·(3+2f 0 (1)) = f 0 (1+1)(3+2) = 5f 0 (2) = 5·2 = 10 so y = 2 is the vertical asymptote (for both +∞ and −∞). 1 The −2 −2/3 f (1) = 1, f (2) = 2, f 0 (1) = 1 f 0 (2) = 2 The horizontal asymptotes: x→±∞ −1 −1 14. Suppose f is differentiable so that: 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). 2x2 2 = lim 2 x −x−2 x→±∞ 1 − 1 − x 0 0 The minimum occurs at x = −1 and the maximum occurs at x = 0. The minimum value is −1 and the maximum value is 0. 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. 1 x→1 (x2 + x + 1) − x(2x + 1) −x2 + 1 = 2 2 2 (x + x + 1) (x + x + 1)2 so the critical points are x = ±1 of which we are only concerned with x = −1. Now build a chart of values: ln(x) 1−x x→1 lim 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. x1000 − 1 1000x999 = lim = 1000 lim x−1 1 x→1 x→1 x→0 x→2+ lim f (x) = lim mx + b = 2m + b x→1 (i) lim x≤2 x>2 • To compute the limit, we have to do them separately: lim f (x) = lim x2 = 4 2xex2 −1 x−1 Using L’Hospital’s rule: (h) lim if if • f (2) = 2m + b, so f (2) exists. 3x2 = lim ex2 x2 mx + b 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: x→∞ x3 n function is f (x) = x−2 at x = 1 15. Let x2 y + a2 xy + λy 2 = 0 4 (a) Let a and λ be constants, and let y be a function of dy x. Calculate dx : 2xy + x2 (e) If y = ax + b, then dy dy −(2xy + a2 y) = −(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 dy =x+0=x da (f) 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. (g) If f 0 (r) exists, then dλ ? da lim f (x) = f (r) 16. 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). x→r 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. (h) 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) (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. (j) (b) If f is differentiable, then limπ θ→ 3 1 2 cos(θ) − = − sin π 3 θ− lim True, since x→a = 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 : (l) sin−1 sin

2π 3 =  sin−1 sin (d) If f 0 (x) exists and is nonzero for all x, then f (1) 6= f (0). True. If f 0 (x) exists for all x, then f is differentiable everywhere (and is also continuous everywhere). Thus, the Mean Value Theorem applies. If f (1) = f (0), that would imply the existence of a c in the interval (0, 1) so that  2π 3  = π 3 (m) 5log5 (2x) = 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. (n) f (1) − f (0) =0 1−0 d ln(|x|) = dx True: 1 , x for all x 6= 0. ln |x| = n ln(x), ln(−x), if x > 0 if x < 0 so but we’re told that f 0 (x) 6= 0. = 2π 3 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, d (10x ) = 10x ln(10) dx dλ da f (x) − f (a) x−a (k) 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. d (10x ) dx answer is π 3   with f (x) = cos(x) and a = π3 . This is another way of defining the derivative of cos(x) at x = π3 . 1 f 0 (x) d p f (x) = (f (x))−1/2 · f 0 (x) = p 2 dx 2 f (x) 2 The = True by l’Hospital’s rule. You could have also said that the expression is in the form: d p f 0 (x) f (x) = p dx 2 f (x) f 0 (c) = d (f (g(x)) dx (i) 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). 17. True or False, and give a short explanation. (c) =x dy True. If we’re computing da , then we’re treating x and b as constants. Differentiating, we get dy dy dy + a2 y + a2 x + 2λy =0 dx dx dx (x2 +a2 x+2λy) dy da d ln |x| = dx 2axy −y 2 5  1 , x 1 −x · (−1) = 1 , x if x > 0 if x < 0 (o) d 10x dx = x10x−1 Oops- a duplication. See part (c) 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 (p) 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. (q) False. The most general antiderivative is a piecewise defined function, since x = 0 is not in the domain of f:  −1 + C1 , if x<0 x f (x) = −1 + C2 , if x>0 x 23. For what value(s) of c does f (x) = cx4 − 2x2 + 1 have both a local maximum and a local minimum? First, f 0 (x) = 4cx3 − 4x = 4x(cx2 − 1), and f 00 (x) = 12cx2 − 4. The candidates for the location of the local 0 max’s and p min’s are where f (x) = 0, which are x = 0 and x = ± 1/c (c > 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 18. Find the domain of ln(x − x2 ): Use a sign chart to determine where x−x2 = x(1−x) > 0: x 1−x − + x<0 + + 01 p are local mins at x = ± 1/c. √ 24. 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 lim ·p √ h h→0 1 − 2(x + h) + 1 − 2x so overall, x − x2 > 0 if 0 < x < 1. 19. 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 1 2 x = −2. We should get that f 0 (x) = (x+2) 2 , f (1) = 3 and f (4) = 23 . Therefore, the Mean Value Theorem says that c should satisfy: 2 1 −3 2 1 = 3 = (c + 2)2 4−1 9 = lim h→0 or √ (c + 2)2 = 18 ⇒ c = −2 ± 18 √ of which only −2 + 18 ≈ 2.243 is inside our interval. lim h→0 1 − 2(x + h) + √ = √ 1 − 2x −1 1 − 2x 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. ...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: 21. 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? 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: y 00 = 6x + 2a 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: y − 8 + a2 = −2a(x − a) y = x3 − 3x2 + bx + 1 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: so we take the leftmost value, a = −2 p 25. 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). 20. 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. −8 + a2 = −2a(4 − a) = −8a + 2a2 ⇒ √ 8 ± 32 ⇒ a= 2 1 − 2x − 2h − 1 + 2x = √ h( 1 − 2(x + h) + 1 − 2x) p and we want the curve to go through the point (1, 6): 6=1−3+b+1 0 = a2 − 8a + 8 so b = 7. 26. 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: √ 8− 32 . 2 F 0 (3) = f 0 (g(3))g 0 (3) 22. Compute the limit, without using L’Hospital’s Rule. √ x+2−3 lim x−7 x→7 so F 0 (3) = f 0 (6) · 4 = 7 · 4 = 28 6 33. If f (x) = (2 − 3x)−1/2 , find f (0), f 0 (0), f 00 (0). Differentiate: 27. 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: f 0 (x) = 3 −1 (2 − 3x)−3/2 (−3) = (2 − 3x)−3/2 2 2 −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) = 1 f (0) = √ , 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: f 0 (0) = 8 1 2x = √ , y = 5 3 6 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 2z = 2A + 2B dt dt dt The two’s divide out and put in the numbers: 29. 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: 00 2 f (x) = 4t − 6t , 0.5 · and solve for 2 f (t) = 12t − 12t = 0 ⇒ t = 0, t = 1 dy = −2x dx = −2 · 2 · (0.4) = 1.6 √ 36. Find the linearization of f (x) = 1 − x at x = 0. To linearize, we find the equation of the tangent line. f 0 (x) = 1 (1 − x)−1/2 (−1) 2 so f 0 (0) = − 12 , and the point is (0, 1). 1 1 y − 1 = − x, or y = − + 1 2 2 √ 37. 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 32. Find a function f (x) so that f 0 (x) = 4 − 3(1 + x2 )−1 , and f (1) = 0 Antidifferentiate: Now, f 0 (t) = (x) + C f (t) = and f (1) = 0 means: 4 − 3 tan−1 (1) + C = 0 tan−1 (1) dz , 78. dt ∆y = f (−2 + 0.4) − f (−2) = 3.44 − 2 = 1.44 P (x) = x2 − x + 3 f (x) = 4x − 3 tan dz = 0.3 · 50 + 0.4 · 60 dt 35. Compute ∆y and dy for the value of x and ∆x: f (x) = 6 − x2 , x = −2, ∆x = 0.4. Recall that ∆y = f (x + ∆x) − f (x) and ∆x = dx. Also, dy = f 0 (x) dx. Put the numbers in: The velocity at t = 0 is 0 and the velocity at t = 1 is 4 − 6 = −2. √ 30. If y = 5t − 1, compute y 000 . Nothing tricky here- Just differentiate, and differentiate, and differentiate! 5 1 y 0 = (5t − 1)−1/2 · 5 = (5t − 1)−1/2 2 2 5 −1 −25 (5t − 1)−3/2 · 5 = (5t − 1)−3/2 y 00 = · 2 2 4 375 y 000 = (5t − 1)−5/2 8 31. Find a second degree polynomial so that P (2) = 5, P 0 (2) = 3, and P 00 (2) = 2. The general form of a second degree polynomial is ax2 + bx+c, so we need to find a, b, c. So let P (x) = ax2 +bx+c. Then P 0 (x) = 2ax + b, and P 00 (x) = 2a. P 00 (2) = 2, so 2a = 2 and a = 1. Now, P 0 (x) = 2x + b, and P 0 (2) = 3, so 3 = 4 + b, and b = −1. Finally, P (x) = x2 − x + c and P (2) = 5, so 5 = 4 − 2 + c, so c = 3. −1 27 √ 16 2 z 2 = A2 + B 2 √ 28. 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. 3 f 00 (0) = 34. 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: 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: 0 3 1 3 · = √ , 2 23/2 4 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: π 4 1 4 5 5 + 8 + 25 19 −4 + + +C = 1 ⇒ +C = 1 ⇒ C = 1− = 6 15 6 30 15 15 Now, = (it’s where sin(x) = cos(x)), so C = 3π − 4. so f (x) = 4x − 3 tan−1 (x) + 3π −4 4 4 7 38. Find f 0 (x) directly from the definition of the derivative (using limits and without L’Hospital’s rule): First, recall that: 43. 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 = f (x + h) − f (x) f (x) = lim h h→0 0 We’ll use this for these exercises: √ (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
√ √ 3 − 5x − 5h + 3 − 5x lim h→0 h (a) f (x) = 4 − x2 + 3ex F (x) = 4x − 13 x3 + 3ex + C (b) f (x) = (b) f (x) = x2 (c) f (x) = (x + h)2 − x2 x2 + 2xh + h2 − x2 lim = lim = h h h→0 h→0 1 x = lim h h→0 lim − h→0 x−(x+h) x(x+h) h 2x + = 6+ 39. If f (0) = 0, and = 2, find the derivative of f (f (f (f (x)))) at x = 0. First, note that the derivative is (Chain Rule): f 0 (0) · f 0 (0) · f 0 (0) · f 0 (0) = 24 = 16 f (x) = if x ≥ 0 if x < 0 f 0 (x) = 41. f (x) = | ln(x)|. Find f 0 (x). We can rewrite f (Recall that ln(x) < 0 if 0 < x < 1) if x ≥ 1 if 0 < x < 1 ln(x), − ln(x), x2 − x −x2 + x if x ≤ 0, or x ≥ 1 if 0 < x < 1 n 2x − 1 −2x + 1 if x < 0, or x > 1 if 0 < x < 1 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). n 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 , so f 0 (x) → ∞ as x → 0+ then f 0 (x) = 2√ x f (x) = −32 (x − 3) 35 47. 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: which simplifies (since f (0) = 0) to: f (x) = 3 0 −32 y + 2 + 8y 0 = 0 ⇒ y 0 = 4 35 y−4= f 0 (f (f (f (x)))) · f 0 (f (f (x))) · f 0 (f (x)) · f 0 (x) √ x √ − x 1 −1/2 0 √ y y x + y + 2yy 0 = 0 2 At x = 3, y = 4: f 0 (0) n 2 3/2 x +C 3 46. Find the slope of the tangent line to the following at the √ point (3,4): x2 + yx + y 2 = 31 −h −1 −1 = lim = 2 hx(x + h) x h→0 x(x + h) 40. Differentiate: + 1 F (x) = −3x−1 + 2 ln |x| + x + C F (x) = 2x1/2 + (c) f (x) = x−1 1 x+h 3 + x2 x2 1+x √ x First rewrite f (x) = x−1/2 + x1/2 , and h(2x + h) lim = 2x h h→0 h→0 44. 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. 45. Find the general antiderivative: −5 √ 2 3 − 5x lim −(2x + y) x + 3y 2 48. Does there exist a function f so that f (0) = −1, f (2) = 4, and f 0 (x) ≤ 2 for all x? and differentiate piecewise: f 0 (x) =  1 , x − x1 , f 0 (x) = if x > 1 if 0 < x < 1 Since 52 > 2, there can exist no function like that (that is continuous). √ 49. Linearize f (x) = 1 + x at x = 0. Point: x = 0, y = 1 Slope: f 0 (0) = 12 Note that the pieces don’t match at x = 1; we remove that point from the domain. 42. f (x) = xeg( √ x) . Find f 0 (x). f 0 (x) = eg( √ x) + xeg( √ x) f (2) − f (0) 5 = 2−0 2 √ 1 · g 0 ( x) · x−1/2 2 Line: y − 1 = 8 1 (x 2 − 0), or y = 1 x 2 +1 √ 50. Find dy if y = 1 − x and evaluate dy if x = 0 and dx = 0.02. Compare your answer to ∆y 55. Sketch the graph of a function that satisfies all of the given conditions: f (1) = 5 lim f (x) = ∞, −1 1 dy = √ dx, ⇒ dy = √ · 0.02 = 0.01 2 1−x 2 1−0 √ √ ∆y = 1 − 0.02 − 1 = 0.01005... dt and ds . 56. If s2 t + t3 = 1, find ds dt First, treat t as the function, s as the variable: 2st + s2 −(s2 + 3t2 ) ds = dt 2st which you can either state directly or show. 3π f (x) = 1 + sin(x) = 0 ⇒ sin(x) = −1 ⇒ x = 2 0 57. Antidifferentiate: √ (a) f 0 (x) = 3 x − is the only critical point in [0, 2π]. Now the table: so that f (x) = 3 · (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 2 = 2 + 2 + c ⇒ c = −2 Finally, f (x) = 2x3/2 + 2x1/2 − 2 (b) f 00 (x) = x2 + 3 cos(x), 2π 1 so f is increasing most rapidly at x = 2 · x3/2 + 2x1/2 + C 3 Now, f (1) = 2, so: f (0) = 2, f 0 (0) = 3 1 3 x + 3 sin(x) + C 3 with f 0 (0) = 3, so C = 3. Now, f 0 (x) = π . 2 (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 0 (x) = 1 3 x + 3 sin(x) + 3 3 so that f (x) = 53. 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. 1 1 4 1 4 · ·x −3 cos(x)+3t+C = x −3 cos(x)+3t+C 3 4 12 and f (0) = 2, so 2 = −3 + C ⇒ C = 5 so that: f (x) = 1 4 x − 3 cos(x) + 3t + 5 12 (c) f 00 (x) = 3ex + 5 sin(x), f (0) = 1, f 0 (0) = 2 We’ll do this one a bit differently, although we could do it like before: f 0 (x) = 3ex − 5 cos(x) + C1 54. Approximate the change in volume of a cone, if we assume the height to be constant and r changes from 2 to 2.1. (V = 13 πr2 h) Approximation means to use dV , which is (h is constant:) dV = f (1) = 2 f 0 (x) = 3x1/2 − x−1/2 so f is greatest at x = 2π, least at 0. x f 0 (x) √1 , x First, rewrite: 2π 2π − 1 ≈ 5.2 ≈ 4.7 dt dt dt −2st + 3t2 =0⇒ = 2 ds ds ds s + 3t2 For s as the function, t as the variable: (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, 3π 2 3 2 π f 0 (1) = f 0 (4) = 0 f (2) = 4 (We’ll do this one in class) 52. If f (x) = x − cos(x), x is in [0, 2π], then find the value(s) of x for which 0 −1 x→2− x→2+ 51. Fill in the question marks: If f 00 is positive on an interval, then f 0 is INCREASING and f is CONCAVE UP. x f (x) f (4) = 2 lim f (x) = 3 f (x) = 3ex − 5 sin(x) + C1 t + C2 To find C1 , C2 , plug in the numbers: From f 0 (0) = 2, 2 = 3 − 5 + C1 ⇒ C1 = 4 2 πrh dr 3 To find C2 , we do something similar: 1 = 3 + 0 + 4 · 0 + C2 ⇒ C2 = −2 Plug in the numbers, using r = 2 and dr = 0.1 to get so: f (x) = 3ex − 5 sin(x) + 4t − 2 dV ≈ 0.133πh 9 (d) f 0 (x) = √ 4 , f (1/2) 1−x2 =1 We see that: f 0 (x) = 4 · √ 1 1 − x2 so that: f (x) = 4 sin−1 (x) + C Now, at x = 1/2, 1 = 4 sin−1 (1/2) + C ⇒ 1 = 4 · and f (x) = 4 sin−1 (x) + 1 − π 2π +C ⇒ C = 1− 6 3 2π 3
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