   Chapter 4.4, Problem 8E Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Solutions

Chapter
Section Calculus: Early Transcendental Fun...

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

Determining Concavity In Exercises 5-14, determine the open intervals on which the graph of the function is concave upward or concave downward. f ( x ) = 2 x 2 3 x 2 + 1

To determine

To calculate: The open intervals for which the provided function f(x)=2x23x2+1 is concave upwards or concave downwards.

Explanation

Given:

The function f(x)=2x23x2+1.

Formula used:

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:

The provided function is continuous on the whole real number line.

Now differentiate the provided function twice.

f'(x)=4x(3x2+1)6x(2x2)(3x2+1)2=4x(3x2+1)2f''(x)=(4(3x2+1)24x(2(3x2+1)2(6x))((3x2+1)2)2)=4(9x21)(3x2+1)3

Now equate the second derivative to zero to obtain the inflection point.

4(9x21)(3x2+1)3=09x21=09x2=1x=13,13

This gives rise to three intervals (,13),(13,13),(13,)

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