`F(x) = xsqrt(6 - x)` (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...

`F(x) = xsqrt(6 - x)` (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 39 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=xsqrt(6-x)` 

differentiating,

`f'(x)=xd/dxsqrt(6-x)+sqrt(6-x)`

`f'(x)=x(1/2)(6-x)^(-1/2)(-1)+sqrt(6-x)`

`f'(x)=-x/(2sqrt(6-x))+sqrt(6-x)`

`f'(x)=(-x+2(6-x))/(2sqrt(6-x))`

`f'(x)=(-3x+12)/(2sqrt(6-x))`

`f'(x)=(-3(x-4))/(2sqrt(6-x))`

Lt us find the critical number by setting f'(x)=0 

`(-3(x-4))/(2sqrt(6-x))=0`

x=4 

Domain of the the function : x `<=` 6` `

Now let us find out the sign of f'(x) at test point in the interval (-`oo` ,4) and (4,6)

`f'(3)=(-3(3-4))/(2sqrt(6-3))=sqrt(3)/2`

`f'(5)=(-3(5-4))/(2sqrt(6-5))=-3/2`

Since f'(3) is positive , function is increasing in the interval (-`oo` ,4)

f'(5) is negative so function is decreasing in the interval 4`<x<=6` 

Local maximum can be found by plugging in the critical number x=4 in the function.

`f(4)=4sqrt(6-4)=4sqrt(2)`

Local maximum = 4`sqrt(2)`  at x=4

Now to find the intervals of concavity and inflection points , let us find out the second derivative,

`f''(x)=-3/2((sqrt(6-x)-(x-4)(1/2)(6-x)^(-1/2)(-1))/(6-x))`

`f''(x)=-3/2((sqrt(6-x)+(x-4)/(2sqrt(6-x)))/(6-x))`

`f''(x)=-3/2((2(6-x)+x-4)/(2(6-x)^(3/2)))`

`f''(x)=-3/4((12-2x+x-4)/(6-x)^(3/2))` 

`f''(x)=3/4((x-8)/(6-x)^(3/2))`

Now let us set f''(x)=0 for determining the inflection point and intervals of concavity.

`3/4((x-8)/(6-x)^(3/2))=0 , x=8`

We have to consider the value of x in the domain of the function

So check the concavity of the function by plugging test point in the interval

-`oo<x<=6`

`f''(5)=3/4((5-8)/(6-5)^(3/2))=-9/4` 

Since f''(5) is negative so the function is concave down in the interval

-`oo<x<=6` and there is no inflection point as there is no change in concavity.

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