   Chapter 4.4, Problem 31E 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

Finding Points of Inflection In Exercises15-36, find the points of inflection and discuss the concavity of the graph of the function. y = e − 3 / x

To determine

To calculate: The points of inflection for the provided function y=e3x and thus its concavity.

Explanation

Given:

The function y=e3x.

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 I n 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 this twice,

y'(x)=e3x(3x2)f''(x)=e3x(3x2)(3x2)+e3x(6x3)=(96x)e3xx4

Now equate the second derivative to zero to obtain,

(96x)e3xx4=096x=06x=9x=32

Also, the second derivative does not exist when x takes the value 0.

Substitute this value in the function to obtain,

y(32)=e3(32)=e2

The inflection point divides this into three intervals (,0),(0,32),(32,)

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