Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 39

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 4Local maximum can be found by plugging in the critical number x=4 in the function.
f(4)=4sqrt(6-4)=4sqrt(2)
Local maximum = 4sqrt(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
-oof''(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