Chapter 4.4, Problem 20E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Finding Points of Inflection In Exercises15-36, find the points of inflection and discuss the concavity of the graph of the function. f ( x ) = ( x − 2 ) 3 ( x − 1 )

To determine

To calculate: The points of inflection for the provided function f(x)=(x2)3(x1) and thus its concavity.

Explanation

Given:

The function f(x)=(xâˆ’2)3(xâˆ’1).

Formula used:

The inflection point of the function corresponds to the point where its second derivative disappears.

For a function f that is twice differentiable on an open interval I:

If f''(x)>0 for all x in I, the function f is concave upwards on I and if f''(x)<0 for all x in I, the function f is concave downwards on I.

Calculation:

Now differentiate the provided function twice.

f'(x)=3(xâˆ’2)2(xâˆ’1)+(xâˆ’2)3f''(x)=6(xâˆ’2)(xâˆ’1)+3(xâˆ’2)2+3(xâˆ’2)2=6(xâˆ’2)(xâˆ’1)+6(xâˆ’2)2=6(xâˆ’2)(2xâˆ’3)

Equate the second derivative to zero to obtain the inflection point.

6(xâˆ’2)(2xâˆ’3)=0x=32,2

Now substitute this for x in the function to obtain,

f(32)=(32âˆ’2)3(32âˆ’1)=âˆ’116

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