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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

If f ( x ) = cos 2 x x 2 + x + 1 , π x π , use the graphs of f, f and f to estimate the x-coordinates of the maximum and minimum points and inflection points of f.

To determine

To estimate:

The x – coordinates of the maximum and minimum point and inflection point of f.

Explanation

1) Concept:

i) First derivative test: f' changes from positive to negative at c  then c  is local maximum, if f'  changes from negative to positive at c  then c  is local minimum. If f'  is positive or negative on both left and right side of c  then c  hais no local maximum or minimum.

ii) Inflection point occurs when concavity changes from upward to downward or from downward to upward.

a) Concave upward: f''x>0 ;I  then graph of f is concave upward on I.

b) Concave downward: f''x<0 ;I  then graph of f is concave downward on I

2) Given:

fx=(cosx)2(x2+x+1 , -πxπ

3) Calculation:

Consider the given function,

 fx=(cosx)2(x2+x+1 , -πxπ 

First we find f'x & f''x

To find f'x  differentiate f(x) with respect to x,

d(fx)dx=ddxcosx2(x2+x+1 

Use the quotient rule of differentiation,

f'x=ddxcosx2·(x2+x+1 -ddx(x2+x+1 ·cosx2((x2+x+1 ) 2 

Differentiating, it becomes

f'x=2cosx·(-sinx)·(x2+x+1 -1·ddxx2+x+12·(x2+x+1 ·cosx2x2+x+1

Use power rule of differentiation,

f'x=-2cosx·sinx·(x2+x+1 -(2x+1)·cosx22·(x2+x+1 x2+x+1

Factor out -cosx  from the numerator, and multiply numerator and denominator by 2(x2+x+1 ,

f'x=(-cosx)4sinx·x2+x+1+(2x+1)·cosx[2·x2+x+132 ]

To find f''x  differentiate f'x  with respect to x,

Substitute sin2x=2sinx·cosx

f'x=-2sin2xx2+x+1+cos2x(2x+1)[2·x2+x+132 ] 

 ddx·f'x=ddx-2sin2x·x2+x+1+(2x+1)cos2x[2x2+x+132 ]

Use quotient rule of differentiation,

f''x=-[ddx·2sin2x·x2+x+1+(2x+1)cos2x·[2x2+x+132 ]-ddx·[2x2+x+132 ]·2sin2x·x2+x+1+(2x+1)cos2x][2x2+x+132 ]2

To find,f''x find ddx·2sin2x·x2+x+1+(2x+1)cos2x

Using product rule of differentiation,

ddx·2sin2x·x2+x+1+(2x+1)cos2x=x2+x+1ddx·2sin2x+2sin2x·ddxx2+x+1+2x+1ddxcos2x+cos2x·ddx(2x+1)

Differentiating, it becomes

=x2+x+1·2·cos2x·2+2sin2

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