   Chapter 3.R, Problem 29E

Chapter
Section
Textbook Problem

# 29-32 Produce graphs of f that reveal all the important aspects of the curve. Use graphs of f and f ″ estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points. In Exercise 29 use calculus to find these quantities exactly. f ( x ) = x 2 − 1 x 3

To determine

To produce:

Graph of the given function,fx=x2-1x3

Explanation

1) Concept:

i. Intervals of increase or decrease-.

Compute f' and use I/D test ( if f>0 then f is increasing, if f< 0 then f is decreasing) to find intervals on which f(x) is positive or negative

ii. Find critical numbers. ( the number c such that f(c) = 0 or f(c) does not exist)

iii. Local maximum or minimum values-

Use first derivative test (f' changes from positive to negative at c then c is the local maximum, if f' changes from negative to positive at c then c is the local minimum. If f' is positive or negative on both the left and right side of c, then c has no local maximum or minimum).

iv. Also, we can use the second derivative test (if f (c) = 0, f(c) >0 then c is the local minimum point. f(c) <0 then c is the local maximum point.)

v. Compute f(x) and use the concavity test (the curve is concave upward if f>0 and concave downward if f<0). Inflection points occur when the direction of concavity changes.

2) Given:

fx=x2-1x3

3) Calculation:

fx=x2-1x3

fx=0x=±1

Differentiate f with respect to x,

f' x=x32x-x2-13x2x6

Simplifying,

=3-x2x4

Graph of f(x)

Using I/D Test, and from the graph, we find the following values

As, f'>0 on -1.732, 0,(0, 1.732)

Therefore, interval of increase for fx:-1.732, 0,(0, 1.732)

f'<0 on (-, -1.732)and (1.732, )

Therefore, interval of decrease for fx: (-, -1.732)and (1.732, )

Again, differentiate both sides with respect to x,

f''x=x4-2x-3-x24x3x8

Simplifying,

=2x2-12x5

Graph of f(x)

Using the first and second derivative tests and graphing of f and f, we obtain the following values

As, f''>0 on -2.449, 0, (2.449, )  and

f''<0 on -, -2.449, (0, 2.449)

Therefore, function f is

Concave upward on: -2.449, 0, (2.449, )

Concave downward on: -, -2.449, (0, 2.449)

Evaluating f(x) at ± 2.449

Inflection points: -2.449, -0.34, (2.449, 0.34)

f  has local maximum value about f1

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