1) Find the derivative of f(x) = 2 divided by x at x = -2. (6 points)

a) 1

b) 1/2

c) -1/2

d) -1

2)Find the derivative of f(x) = negative 3 divided by x at x = -4. (6 points)

a) 3/4

b)3/16

c)16/3

d)4/3

3)Find the indicated limit, if it exists. (7 points)
limit of f of x as x approaches -4 where f(x)= x + 3 when x < -4 and f(x) = 3 - x when x is greater than or equal to -4

a)The limit does not exist.

b)-1

c)-4

d)7

4) Find the indicated limit, if it exists. (7 points)
limit of f(x) as x approaches 0 where f(x)= 9 - x squared when x < than 0, 9 when x= 0, and -4x + 9 when x > 0

a) the limit does not exist.

b)9

c)-4

d) -13

5) Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 2 as x approaches 2 from the left . (7 points)

a) ∞; x = -2

b) -∞; x = -2

c) 1 ; no vertical asymptotes

d) -∞; x = 2

6) FREE RESPONSE
The position of an object at time t is given by s(t) = -9 - 3t. Find the instantaneous velocity at t = 8 by finding the derivative. (9 points)

7) FREE RESPONSE
Use graphs and tables to find the limit and identify any vertical asymptotes of the function. (8 points)
limit of 1 divided by the quantity x minus 2 squared as x approaches 2

Problems 1 and 2 involve f(x) of the form f(x) = c / x = c * x^(-1).

The derivative of f(x) is

f’(x) = -c*x^(-2) so just plug in c and x

1) Find the derivative of f(x) = 2 divided by x at x = -2. (6 points)

c= 2, x = -2

f’(x) = -c*x^(-2) = -2 / (-2)^2 = - 1/2

2)Find the derivative of f(x) = negative 3 divided by x at x = -4. (6 points)

c= -3 , x = -4

f’(x) = -c*x^(-2) = 3 / (-4)^2 = 3/16

3)Find the indicated limit, if it exists. (7 points)

limit of f of x as x approaches -4 where f(x)= x + 3 when x < -4 and f(x) = 3 - x when x is greater than or equal to -4

f(x) = x+ 3 when x < -4

f(x) = 3-x when x >= -4

approaching x=-4 from the left:

Lim x-> -4 of f(x) when x < -4, is -1.

approaching x=-4 from the right:

Lim x-> -4 of f(x) when x >= -4, is 7.

So we have a discontinuity at x= -4 because f(-4) has different values when you approach it from the left and the right. Thus the limit doesn’t exist.

4) Find the indicated limit, if it exists. (7 points)

limit of f(x) as x approaches 0 where f(x)= 9 - x squared when x < than 0, 9 when x= 0, and -4x + 9 when x > 0

f(x) = 9 - x^2, when x<0

f(x) = 9 when x = 0

f(x) = -4x + 9 when x>0

Lim from the left is 9.

f(0) = 9.

Lim from right is 9.

No discontinuity and lim x->0 is 9

5) Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 2 as x approaches 2 from the left . (7 points)

f(x) = 1/(x-2), lim as x -> 2 from left means x gets closer and closer to 2 but it’s always less than 2. For example x gets closer to 2 forever like 1.999, 1.99999, 1.999999 but never equals 2.

So x -2 is always negative. And x-2 gets closer and closer to 0. So 1/(x-2) gets bigger and bigger forever as x gets closer to 2 forever. This is because the denominator gets smaller and smaller so the whole term 1/(x-2) gets bigger and bigger.

So we get a limit of -infinity because remember x-2 is negative as x approaches 2 from the left. In other words the function is decreasing infinitely.

Limit from left is -∞

The asymptote of the function is at x=2 because f(2) is undefined AND because the closer we get to x=2 the function f(x) decreases infinitely (approaching from left)

d) -∞; x = 2

6) FREE RESPONSE

The position of an object at time t is given by s(t) = -9 - 3t. Find the instantaneous velocity at t = 8 by finding the derivative. (9 points)

Velocity is the derivative of position with respect to time.

v(t) = s’(t) = -3, there’s no t in the function after getting the derivative which means the velocity is always -3 no matter the time.

7) FREE RESPONSE

Use graphs and tables to find the limit and identify any vertical asymptotes of the function. (8 points)

limit of 1 divided by the quantity x minus 2 squared as x approaches 2

f(x) = 1/(x-2)^2.

To look for asymptotes you have to first explore where the function is undefined. Aka where there is division by 0.

We see f(x) is undefined

when (x-2) = 0 or x = 2.

Using same reasoning as problem 5 you see that f(x) increases infinitely when approaching x = 2 from both sides. (x-2)^2 gets closer and closer to 0 and is always positive no matter if we approach from left or right... because it’s squared. so f(x) = 1/(x-2)^2 gets bigger and bigger as x approaches 2 from either side.

So asymptote at x=2 and

lim f(x) as x->2 is infinity from left and right side.

6 equilateral triangles