   Chapter 3.R, Problem 31E

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 ) = 3 x 6 − 5 x 5 + x 4 − 5 x 3 − 2 x 2 + 2

To determine

To produce:

Graph of the given function

fx=3x6-5x5+x4-5x3-2x2+2

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 the 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 the first derivative test (f' changes from positive to negative at c then c is local maximum, and if f' changes from negative to positive at c then c is local minimum. If f' is positive or negative on both the left and right sides 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=3x6-5x5+x4-5x3-2x2+2

3) Calculation:

We have,

fx=3x6-5x5+x4-5x3-2x2+2

Differentiate with respect to x

f' x=18x5-25x4+4x3-15x2-4x

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-0.23, 0and 1.62,

f'<0 on(--0.23)and (0, 1.62)

Therefore f has,

Interval of increase: -0.23, 0 and (1.62, )

Interval of decrease: --0.23 and (0, 1.62)

Again, differentiate both sides with respect to x,

f''x= 90x4-100x3+12x2-30x-4

Now draw graph for f(x),

Graph of f(x)

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