`h(x) = (x + 1)^5 - 5x - 2` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d)...

`h(x) = (x + 1)^5 - 5x - 2` (a) FInd the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the information from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.

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Textbook Question

Chapter 4, 4.3 - Problem 37 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`h(x)=(x+1)^5-5x-2`

differentiating,

`h'(x)=5(x+1)^4-5`

Now let us find the critical numbers,

`5(x+1)^4-5=0`  

`5x(x+2)(x^2+2x+2)=0`

solving above ,

x=0 , x=-2 , x=-1+i , x=-1-i

ignore the points that are complex 

Now let us check the sign of f'(x) in the intervals (-`oo` ,-2) , (-2,0) and (0,`oo` )

Now let us take a test point in each interval

`f'(-3)=5(-3+1)^4-5=75`

`f'(-1)=5(-1+1)^4-5=-5`

`f'(1)=5(1+1)^4-5=75`

so f'(-3) is positive therefore function is increasing in the interval(-`oo` ,-2),

f'(-1) is negative therefore function is decreasing in the interval (-2,0)

f'(1) is positive therefore function is increasing in the interval (0,`oo` )

Now to find the local extrema plug in the critical numbers in the function

`f(-2)=(-2+1)^5-5(-2)-2=7` 

`f(0)=(0+1)^5-5(0)-2=-1`

so Local Maximum=7 at x=-2

Local Minimum=-1 at x=-1

Now to find the intervals of concavity and inflection points let us find the second derivative of the function.

` ``f''(x)=20(x+1)^3`` `

Now let us find the point of inflection,

`20(x+1)^3=0`

solving above, x+1=0 , x=-1

Now let us check the concavity of the function by taking test point in the intervals (-`oo` ,-1) and (-1,`oo` )

`f''(-2)=20(-2+1)^3=-20`

`f''(1)=20(1+1)^3=160`

f''(-2) `<`  0 , so function is concave downward in the interval (-`oo` ,-1)

f''(1) `>`  0 , so function is concave upward in the interval (-1,`oo` )

So the inflection point is at x=-1 as the concavity changes.

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