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MATH 105 PRACTICE MIDTERM 1 KEY

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

then if

f(x)
h(x)   =           -  3g(x)
2

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
f(1)  >  g(1)       and      f(2) < g(2)
then if
h(x) = f(x) - g(x)
h(x)  has a root for some value of x between 1 and 2.

Solution:

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)

Solution

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?

Solution:

-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).

Solution

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.

Then

|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.