   Chapter 3.R, Problem 30E

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 3 + 1 x 6 + 1

To determine

To produce:

Graph of the given function

fx=x3+1x6+1

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

iii. Local maximum or minimum values-

Use the first derivative test (f' changes from positive to negative at c then c is local maximum; if f' changes from negative to positive at c then c is local minimum. If f' is positive or negative on both 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 direction of concavity changes.

2) Given:

fx=x3+1x6+1

3) Calculation:

We have,

fx=x3+1x6+1

fx=0 x=-1

Differentiate with respect to x,

f' x=-3x2x6+2x3-1x6+12

Now draw the graph for f(x)

Graph of f(x)

Using I/D Test and graphing, We find the following approximate values

f'>0 on (-1.34, 0.75)

f'<0 on -, -1.34 and (0.75, )

Therefore f has,

Interval of increase: (-1.34, 0.75)

Interval of decrease: (-, -1.34)and (0.75, )

Again, differentiate both sides with respect to x,

f''x= 6x2x12  +7x9-9x6-5x3+1x6+13

Graph of f(x)

Using the first and second derivative tests and graphs of f andf, we obtain the following approximate values

f''>0 on-1.64, -0.82, 0, 0.54and 1

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