Define Bounded Area Formula

1. Define A(x) to be the area bounded by the t-axis, the line y = 2t and a vertical line at t = x.
(a) Find a formula for A(x).
(b) Determine (x)

Solution:

(a)

(b)

2. The figure below shows the graph of the derivative of a continuous function f .
(a) List the critical numbers of f .
(b) What values of x result in a local maximum?
(c) What values of x result in a local minimum?

Solution:

(a)
Critical numbers are x = 2, 7

(b)
Function has local maximum at x = 7.

(c)
Function has local minimum at x = 2.

3. Use information from the derivative of each function to help you graph the function. Find all local maximums and minimums of each function.

Solution:

To find critical points, put g'(x) = 0

Critical points are x = 0, 5

Since g"(0) < 0 so g(x) will have local maximum at x = 0.
Since g"(5) > 0 so g(x) will have local minimum at x = 5.

Interval Test value Value of f'(x) at test value Sign of f'(x)
(-∞,0) -1
positive
(0,5) 1
negative
(5,∞) 6
Positive

Since value of g'(x) is positive in the intervals (-∞,0) and (5,∞) and value of g'(x) is negative in the interval (0,5). Thus, function is increasing in the intervals (-∞,0) and (5,∞)and decreasing in the interval (0,5).

Find g''(x)

To find inflection points, put g''(x) = 0

12x-30 = 0
12x = 30
x = 30/12 = 5/2

Test value Value of f''(x) at test value Sign of f''(x)
(-∞,5/2) 0
negative
(5/2,∞) 3
Positive

Since g"(x) is negative in the interval (-∞,5/2) so g(x) will be concave down in the interval (-∞,5/2).
Since g"(x) is positive in the interval (5/2,∞) so g(x) will be concave up in the interval (5/2,∞).

In 4 and 5, a function and values of x so that f '(x) = 0 are given. Use the Second Derivative Test to determine whether each point (x, f (x)) is a local maximum, a local minimum or neither.
4.

Solution:

Find the value of h''(x) on x = -2,0,2

Since h"(-2) and h"(2) greater than 0 so h(x) has local minimum at x = -2 and x = 2.

Since h"(0) < 0 so h(x) has local maximum at x = 0.

Thus function has local minimum at (-2, -18) and (2,-18).
Function has local maximum at (0, -2).

5.

Solution:

Find the value of f''(x) on x = 1/e

Since f"(1/e) >0 so f(x) has local minimum at x = 1/e.

Thus function has local minimum at (1/e, -1/e).

6. Lest you have forgotten, the formulas you will need are as follows:
For a right circular cylinder of radius "r" and height "h":
The volume V = πr2h.
The surface area S = circular ends plus the cylindrical wall = 2πr2 + 2πrh.

You have been asked to design a one-liter (i.e., 1000 cm3) can shaped like a right circular cylinder (figure below). What dimensions will use the least material?

So the question is: What should "r" be and what should "h" be such that the volume is 1 liter but the surface area is least?

Solution:

V = πr2h

Given V = 1000 cm3
πr2h = 1000
h = 1000/πr2

We know that

S = 2πr2 + 2πrh
S = 2πr2 + 2πr*1000/πr2

Find S'(r)

Put S'(r) = 0

cm

Thus, S"(r) is positive throughout the domain. Thus, S(r) will have least value at r = 5.42 cm

We know h = 1000/πr2

cm

Answer: r = 5.42 cm and h = 10.84 cm

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