Name
MATH 105 PRACTICE
MIDTERM 1 Please work out each of the given
problems. Credit will be based on
the steps that you show towards the final answer.
Show your work. PROBLEM 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why. A) (7 Points) If f(x) and g(x) are differentiable functions with
f '(5)
= 10 and g'(5)
= 4
f(x) then h'(5) = -7 True: h'(x) = (f(x)/2 - 3g(x))' = f '(x)/2 - 3g '(x) So that h'(5) = f '(5)/2 - 3g '(5) = 10/2 - 3(4) = -7
B)
(7 Points) Let
f(x)
and
g(x)
be continuous functions True: h(1) = f(1) - g(1) > 0 h(2) = f(2) - g(2) < 0 Hence by the Intermediate Value Theorem, there is a c with h(c) = 0. C) (7 Points) Let f(x) and g(x) be continuous functions such that and Then h(x) has a vertical asymptote at x = 2. False, Let f(x) = x - 2 and g(x) = x - 2 PROBLEM 2 Find the following limits if they exist: A) (8 Points)
B) (8 Points)
C) (8 Points)
PROBLEM 3 A) (8 Points) Find the following limits if they exist i) ii) iii) iv) v) i) 0 ii) 2 iii) 1 iv) Does Not Exist v) Does Not Exist
B) (8 Points) At which values is f(x) not continuous? -1, 1, and 3 C) (8
Points) At which values is
f(x) not differentiable? -3, -1, 1, and 3 PROBLEM 4 (20 Points) Below is the function y = f(x). Sketch a graph of the derivative y = f ’(x). PROBLEM 5 Find f ' (x) for the following A) (10 Points)
B) (11 Points) -10x-6 - 2x + 2cosx - 2xsinx + 2cosxsinx + 5/2 x3/2
PROBLEM 6 Let A) (10 Points) Use the limit definition of the derivative to find f ’(x).
B) (10 Points) Prove using the e-d definition of the limit that Let e > 0 , choose d = e/2. |x - 2| < d implies that |x - 2| < e/2 so that |2x - 4| < e or |4 - 2x| < e adding and subtracting two gives |4 + 2 - 2x - 2| < e |6 - 2x - 2| < e Hence |f(x) - 2| < e So that that the limit exists.
PROBLEM 7 (20
Points)
The position of a robin flying through the wind is given by s(t) = -5t + tcost Find its acceleration when t is 2 seconds. The acceleration is just the second derivative, so first compute the first derivative. s '(t) = -5 + cost - tsint Now the second derivative is the derivative of the derivative: s ''(t) = (s'(t))' = -sint - sint - tcost = -2sint - tcost Finally, plug in t = 2 to get s ''(2) = -2sin2 - 2cos2 which is approximately -1. |