Mat 210, Spring 2001 – Ela Jackiewicz

ã 2001 Arizona State University Department of Mathematics

Exam 1 Key – Covers Chapters 2 and 3

1.  Find the following limits. If the limit does not exist clearly explain why.

[4 pts each]

a. = 8 (since e^(-x) -> 0 as x ->+ infinity, check on the graph)

b. = - 17/3

c. =4 ( factor numerator as (x-2)(x+2) and simplify)

d.= 1/4( multiply numerator and denominator by (1/x^3),

expressions like 2/x^2, 5/x^3 will approach 0 as x -> +infinity

e. =

2. Consider the function .

a.  Using the limit definition, find the instantaneous rate of change of at [9pts]

(1/h)[f(3+h)-f(3)]=(1/h) [(3+h)^2-4(3+h) – (3^2-4*3)]= (1/h)[ 9+6h+h^2-12-4h-9+12]

=(1/h)[h^2+2h]=h+2

limit(h+2)=2(as h->0)

b. Using your answer to part a) find the equation of the tangent line at the point [6pts]

f(3)= - 3, from part a) slope = 2, y+3=2(x-3) gives y=2x-9

3.  The graph of is given below.

b) Find . If the limit does not exist clearly explain why. [3pts]

does not exist, since left limit(1) is not the same as right limit (3)

a)  Find. If the limit does not exist clearly explain why. [3pts]

=4

4.  Fill in the table and then estimate

if it exists. [5pts]

x / .9 / .99 / .999 / 1.001 / 1.01 / 1.1
/ .2372 / .2487 / .2498 / .2501 / .2512 / .2623

=.25

5. Find the derivative of the following functions. [5pts each]

a.

b.

c.

d.

6. A particle moves according to the law . Find the velocity at [10pts]

V(t)=s’(t)= v(5) = (68/729)*(e^5) ~ 13.84

7.  The profit function on a certain commodity is given by

, where is the number of items produced

and is the cost in hundred of dollars to produce items.

a. Find the marginal profit function. [4pts]

P’(x)=1.8x^2-6x+12

b.  For find the approximation for using the marginal profit function from part (a.) [3pts]

P’(20) = 612

c.  For , find the profit on the next item, Compare the result to part (b). [3pts]

P(21)-P(20)=645.6

Numbers are relatively close.

8.  Find the points on the graph of where the tangent line to the graph is horizontal. [10pts]

f’(x)= e^x(2x+x^2)=0

x=0 or x=-2

f(0)=0 f(-2)=4e^(-2)

(0,0) and (-2, .54) are the points