Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 11

We need to take the derivative of the function first in order to identify the critical points or critical numbers, which we can use for the endpoints of our intervals.
For the right side we need to apply the Product Rule:
f'(uv) = u'v+v'u
Let us set u = x and v= sqrt(16 - x^2)
Therefore, u' = 1 and v' = 1/2(16-x^2)^(-1/2)*(-2x) = -(x/(16-x^2)^(1/2))
So, we will have:
y' = 1*sqrt(16-x^2) +(-x/(16-x^2)^(1/2))*x
y' = sqrt(16-x^2) - x^2/(16-x^2)^(1/2)
Take note that is the same (16-x^2)^(1/2) as sqrt(16-x^2) .
We will equate it to zero to find the critical numbers.

sqrt(16-x^2) - x^2/sqrt(16-x^2) =0
Multiply both sides by sqrt(16-x^2) .
sqrt(16-x^2)*sqrt(16-x^2) +sqrt(16-x^2)(-x^2/sqrt(16-x^2))=sqrt(16-x^2)*0
Remember that sqrt(a)*sqrt(a) = a for instance, sqrt(3)*sqrt(3) = sqrt(3*3) = sqrt(9)= 3 .
So, we will have:
16 - x^2 - x^2 = 0
Combine like terms.
16 - 2x^2 = 0
Isolate the -2x^2 on left side. So, we subtract both sides by 16.
-2x^2 = -16
Get rid of the -2 on left side, so we multiply both sides by -1/2 .
(-1/2)*(-2x^2) = (-1/2)*-16
x^2 = 8
Take the square root of both sides.
x =+- sqrt(8) = +-sqrt(4*2) = +-2sqrt(2)
Therefore, the critical numbers are x = -2sqrt(2),2sqrt(2) .
By the way,our equation contains sqrt(16-x^2) ,the maximum value that we
can use for our x is 4, since we will get an imaginary number for x= 5 and greater than that. Since, sqrt(16 - (5^2)) = sqrt(16 - 25) = sqrt(-9)
Therefore, the intervals for our graph is (-4,-2sqrt(2)),(-2sqrt(2), 2sqrt(2)), (2sqrt(2), 4) .
We can test each interval using the first derivative, if it is increasing or decreasing.
Let us use -3 for the first interval.
sqrt(16- (-3)^2)-((-3^2))/sqrt(16-(-3^2))
= sqrt(16 - 9) - 9/sqrt(9 - 7)
=sqrt(7) - 9/sqrt(7) = (7 - 9)/sqrt(7)
=-2/sqrt(7)
Rationalizing, it will be:
-2/sqrt(7) = (-2sqrt(7))/7
Therefore, for the first interval (-4, -2sqrt(2)) it is decreasing.
For the second interval, let us plug-in 0.
sqrt(16 - 0^2) - 0^2/sqrt(16 - 0^2) = sqrt(16) = 4
Therefore, it is increasing on the second interval.
For the third interval, let us plug-in 3.
sqrt(16-(3^2)) - (3^2)/sqrt(16 - (3^2))
= sqrt(7 ) - 9/sqrt(7) = (-2sqrt(7))/7
It is also negative, so it is also decreasing for the last interval.
Therefore, the open interval for which function are increasing is (-2sqrt(2), 2sqrt(2)) and the intervals where the function is decreasing are
(-4,2sqrt(2)) and (2sqrt(2), 4) .