`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`

...

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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.