Math 128 – Calculus I – Exam #1



Math 128 (Calculus I) Exam #3 - Review Exercises

1. Suppose f(x) = 300 – x3 – 6x2 + 36x .

(a) Find all critical points.

(b) Find the absolute minimum value of f(x) and the

absolute maximum value of f(x) on the interval

[–12, 5].

2. Suppose f(x) = x4 – 16x3.

(a) Identify the interval(s) where f(x) is increasing and the

intervals where f(x) is decreasing.

(b) Identify the interval(s) where f(x) is concave up and

the intervals where f(x) is concave down.

(c) Draw a sketch of the basic shape of the function, but

you do not have to be concerned with vertical scaling.

3. A total of 60 feet of fence is

available to enclose a region

in the shape of the isosceles

triangle displayed on the

right. The length of the base

of the triangle has been

labeled “x”.

(a) Find a formula for length of

each of the other two sides.

(b) Use the Pythagorean formula

to find a formula for the

height of the triangle.

(c) Find a formula for the area of the triangle.

(d) Find the value of x which maximizes the area of the

triangle.

(e) Use the value of x in part (d) to find the maximum area

possible and the length of each of the other two sides

necessary to obtain this maximum area.

4. Use L’Hopital’s Rule when appropriate to find each of

the limits listed.

(a) lim [pic] (b) lim [pic]

x ( 1 x ( 9

(c) lim [pic] (d) lim [pic]

x ( 1 x ( 3

5. Suppose we intend to apply Newton’s method to solve

x2 + 1 = 0.

(a) Why can we not use the starting value x0 = 0?

(b) If we use the starting value x0 = 1, what do we get for

x1 and x2?

(c) Why will Newton’s method never give us an answer

no matter what starting value we use?

6. Do the following exercises from the Practice Exercises

at the end of Chapter 4: #75, 77, 79, 81.

Answers to Review Exercises

1. (a) x = 2, –6

(b) The absolute minimum value is f(–6) = 84.

The absolute maximum value is f(–12) = 732.

2. (a) f is decreasing on (–(, 0) and on (0, 12).

f is increasing on (12, ().

(b) f is concave up on (–(, 0).

f is concave down on (0, 8).

f is concave up on (8, ().

(c)

[pic]

3. (a) [pic] (b) [pic]=[pic]

(c) [pic]=[pic] (d) x = 20 ft

(e) maximum area = [pic]ft2, other lengths = 20 ft

4. (a) 1/9 (b) 1/2 (c) 3/2 (d) 1/2

5. (a) We have to divide by zero which is not possible.

(b) We get x1 = 0 but must divide by zero to get x2 .

(c) x2 + 1 = 0 has no solution, since x2 + 1 is always

positive.

6. (Answers are in the back of the textbook.)

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