   Chapter 4.4, Problem 28E 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. f ( x ) = sin x + cos x ,       [ 0 , 2 π ]

To determine

To calculate: The points of inflection for the provided function f(x)=sinx+cosx and thus its concavity.

Explanation

Given:

The function f(x)=sinx+cosx over the interval [0,2π].

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 this twice,

f'(x)=cosxsinxf''(x)=sinxcosx

Now equate the second derivative to zero to obtain,

sinxcosx=0sinx=cosxx=3π4,7π4

Substitute these values in the function to obtain,

f(3π4)=sin(3π4)+cos(3π4)=1212=0

And,

f(7π4)=sin(7π4)+cos(7π4)=12+12=0

The domain of the function is [0,2π]. The inflection point divides this into three intervals [0,3π4),(3π4,7π4),(7π4,2π]

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