Math 105 Answer Key Exam #2 2013

Math 105 1. Answer Key Exam #2 November 1, 2013 [10 Points] Suppose that f and g are functions, and • lim f (x) = 9 • g(7) = −6 • lim f (x) = 7 • g(x) is continuous at x = 7 • f (3) = −5 • f (x) is continuous at x = 4 x→3 x→4 Evaluate the following quantities and fully justify your answers. Do not just put down a value: (a) f (4) = lim f (x) = 7 The first equality holds because of the assumption of f being continuous x→4 at x = 4. The second equality was given in the assumptions. (b) lim g(x) = g(7) = −6 The first equality holds because of the assumption of g being continuous x→7 at x = 7. The second equality was given in the assumptions. (c) Compute g ◦ f (4) = g(f (4)) = g(7) = −6 using the answer from (a) and the given values. (d) Is f (x) continuous at x = 3? Why or why not? Use math notation. No , f is not continuous at x = 3 because 9 = lim f (x) 6= f (3) = −5. x→3 2. [35 Points] Compute the derivative of each of the following functions. For these problems, you do NOT need to simplify your derivative. √ 5 5 1 1 5 5 1 (a) y = x + x 6 + 5x + 6 + √ = x + x 6 + (5x + 6) 2 + (5x + 6)− 2 . 6 6 5x + 6   3 5 5 −1 1 5 (5x + 6)− 2 (5) + x 6+ √ + − y′ = 6 6 2 2 5x + 6 (b) y =  5 x + 8 3 x 9    x 5 8 1 40 = 9 + − 3 x8 3 x9   √ 5 2 (c) f (x) = x − 2 (3x + x) x      √ 1 5 10 2 ′ 3+ √ + (3x + x) 2x + 3 f (x) = x − 2 x x 2 x y′  (d) y = √
− 1 1 = x2 − 5x + 3 2 x2 − 5x + 3 1 y′ = −
− 3 1 2 x − 5x + 3 2 (2x − 5) 2 (e) y = y′ =   1 + 7x x3 1 + 7x x3 5 5  7 7 (−5) x4  1 − 7 x x4 −5 1 − 7 x −6 4x3 + 7x−8
continued...  − 2      7 3 1 −5 5 1 4 − 4 +7 + x − 7 + 7x x 7 x3 x 2 3 √ 3 2 x + x  (f) f (x) =  2 2  x3 + x 3  f ′ (x) 3.  − 1 √ 2  2 x + x3  =  2 2  3 x3 + x 3 [10 Points] 3  2  x + 3x    2 3    
2 −1 2 √ 2 2 3 √ + 3x − 2 x + x x 3+ 3 3  2 x   2   2 2 x3 + x 3 Find the equation of the tangent line to this curve y = at the point where x = 1. First we need slope. Compute the derivative.
1 y′ = p 1 + 3(x2 + 1)2 (2x) 2 x + (x2 + 1)3 p x + (x2 + 1)3
1 25 1 1 + 3(2)2 (2) = √ (25) = The specific slope at x = 1 is y ′ (1) = p 3 6 2 9 2 1 + (2) √ √ The point is (1, y(1)) = (1, 1 + 23 ) = (1, 9) = (1, 3) Using point-slope form, we find the equation of the tangent line 25 25 25 y − 3 = (x − 1) or y − 3 = x − 6 6 6 or finally 25 25 +3 y = x− 6 6 25 18 25 y = x− + 6 6 6 y= 25 7 x− 6 6 2 7x + 3 . 1 − 5x (a) Compute the derivative of f using the limit definition of the derivative. 4. [15 Points] Consider the function f (x) = 7(x + h) + 3 7x + 3 − f (x + h) − f (x) 1 − 5(x + h) 1 − 5x ′ = lim f (x) = lim h→0 h→0 h h [7x + 7h + 3](1 − 5x) − (7x + 3)[1 − 5x − 5h] (1 − 5(x + h))(1 − 5x) = lim h→0 h   2 7x + 7h + 3 − 35x − 35xh − 15x − 7x + 35x2 + 35xh − 3 + 15x + 15h (1 − 5(x + h))(1 − 5x) = lim h→0 h   22h   1 22h (1 − 5(x + h))(1 − 5x) = lim = lim h→0 (1 − 5(x + h))(1 − 5x) h→0 h h 22 22 = h→0 (1 − 5(x + h))(1 − 5x) (1 − 5x)2 = lim (b) Compute the derivative of f using the Quotient Rule. f ′ (x) = 22 (1 − 5x)(7) − (7x + 3)(−5) 7 − 35x + 35x + 15 = = 2 2 (1 − 5x) (1 − 5x) (1 − 5x)2 (c) Compute the second derivative f ′′ (x). f ′ (x) = 22(1 − 5x)−2 f ′′ (x) = −44(1 − 5x)−3 (−5) = 5. [10 Points] 220 (1 − 5x)3 Find all x-coordinates at which the graph of the function f (x) = (4x + 1)4 (7 − 3x)8 has horizontal tangent lines. f ′ (x) = (4x + 1)4 8(7 − 3x)7 (−3) + (7 − 3x)8 4(4x + 1)3 (4) = 4(4x + 1)3 (7 − 3x)7 [(−6)(4x + 1) + 4(7 − 3x)] = 4(4x + 1)3 (7 − 3x)7 [−24x − 6 + 28 − 12x] set = 4(4x + 1)3 (7 − 3x)7 [−36x + 22] = 0 This means that which implies 4x + 1 = 0 or x=− 1 4 or 7 − 3x = 0 or x= 7 3 or x= 3 −36x + 22 = 0 11 18 6. [10 Points] Find the equation of the line tangent to the curve x3 + x2 y = 6 − 4y 2 at the point (1, 1). First, implicitly differentiate by taking the derivative of both sides:

d d x3 + x2 y = 6 − 4y 2 dx dx dy dy 3x2 + x2 + y(2x) = 0 − 8y dx dx We can plug the point (1, 1) immediately and solve for the derivative value dy dx dy dy + (1)(2(1)) = 0 − 8(1) dx dx dy dy 3+ + 2 = −8 dx dx dy Isolate and solve for dx dy 9 = −5 dx dy 5 Finally, =− . dx 9 3(1)2 + (1)2 Now the equation of the tangent line at the point (1, 1) with slope − 5 y − 1 = − (x − 1) 9 5 5 or y − 1 = − x − 9 9 5 5 or y = − x − + 1 9 9 5 9 5 or y = − x + + 9 9 9 5 14 or y = − x + 9 9 4 5 is given by 9 7. [10 Points] For each of the following functions, compute the derivative of f and simplify your derivative as much as possible into a single fraction. (a) f (x) = √ x . +1 x2    √ 1 x2 2 √ (2x) + 1 (1) − x x +1− √ 2 x2 + 1 x2 + 1 = f ′ (x) = x2 + 1 x2 + 1 ! √   x2 x2 + 1 √ 2 √ x +1− √ x2 + 1 x2 + 1 = common denominator x2 + 1 √ x2  x2 + 1 − x2 1   √ √ 2 2 1 1 1 x +1 x +1 = = = 2 =√ 3 2 2 2 x +1 x +1 x +1 x +1 (x2 + 1) 2 3 x2 + 1 (b) f (x) = x2 − 1 2  2   2 (x − 1)(2x) − (x2 + 1)(2x) x +1 ′ f (x) = 3 x2 − 1 (x2 − 1)2   2   2  3  2 (x + 1)2 −4x 2x − 2x − 2x3 − 2x x +1 =3 =3 x2 − 1 (x2 − 1)2 (x2 − 1)2 (x2 − 1)2  = −12x(x2 + 1)2 (x2 − 1)4 5 *************************************************************************** OPTIONAL BONUS Do not attempt this unless you are completely done with the rest of the exam. *************************************************************************** OPTIONAL BONUS #1 Compute lim x→0 |x − 1| − |x + 1| − |x| |x| + |2 − x| − |x + 2| DOES NOT EXIST, RHL 6= LHL RHL: lim x→0 = lim x→0+ LHL: lim x→0 |x − 1| − |x + 1| − |x| = lim |x| + |2 − x| − |x + 2| x→0+ −(x − 1) − (x + 1) − x x + (2 − x) − (x + 2) |x − 1| − |x + 1| − |x| = lim |x| + |2 − x| − |x + 2| x→0− −(x − 1) − (x + 1) − (−x) = −x + (2 − x) − (x + 2) −x + 1 − x − 1 − x −3x = lim = 3 + x+2−x−x−2 −x x→0 −x 1 −x + 1 − x − 1 + x = lim = −x + 2 − x − x − 2 3 x→0− −3x  if x ≥ 0  x Here, recall that |x| =  −x if x < 0   if x − 1 ≥ 0  x−1  x−1 |x − 1| = =   −(x − 1) if x − 1 < 0 1−x   if x + 1 ≥ 0  x+1  x+1 |x + 1| = =   −(x + 1) if x + 1 < 0 −x − 1   if 2 − x ≥ 0  2−x  2−x |2 − x| = =   −(2 − x) if 2 − x < 0 x−2   if x + 2 ≥ 0  x+2  x+2 |x + 2| = =   −(x + 2) if x + 2 < 0 −x − 2 = lim x→0− 6 if x≥1 if x<1 if x ≥ −1 if x < −1 if x≤2 if x>2 if x ≥ −2 if x < −2