How do you find the critical numbers for $y = x * sqrt( 9 - x^2)$ to determine the maximum and minimum?

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Answer 1 · 2018-04-01T04:03:22

There is a local minimum at $x=-(3sqrt(2))/2$ , and a local maximum at $x=(3sqrt(2))/2$ .

This equation is in factored form, so it is easy to list the $x$ and $y$ intercepts for this graph.

The $x$ -intercepts are $x=0$ , $x=-3$ , and $x=3$ . The $y$ -intercept occurs at $y=0$ .

Also note that $-3<=x<=3$ .

Now let's take the derivative and find it's intercepts.

$(dy)/(dx)=sqrt(9-x^2)-x^2/(sqrt(9-x^2))=(9-2x^2)/sqrt(9-x^2)$

To find the roots of the derivative, set

$(9-2x^2)/sqrt(9-x^2)=0$ .

This only happens when

$2x^2=9$

$x=+-(3sqrt(2))/2$

When $-3<=x<=-(3sqrt(2))/2$ , $y$ is decreasing because $(dy)/(dx)<=0$ .

When $-(3sqrt(2))/2<=x<=3sqrt(2)/2$ , $y$ is increasing because $(dy)/(dx)>=0$ .

This means that there is a local minimum at $x=-(3sqrt(2))/2$ .

When $(3sqrt(2))/2<=x<=3$ , $y$ is decreasing because $(dy)/(dx)<=0$ .

This means that there is a local maximum at $x=(3sqrt(2))/2$ .

graph{xsqrt(9-x^2) [-4, 4, -5, 5]}

How do you find the critical numbers for y = x * sqrt( 9 - x^2)y = x * sqrt( 9 - x^2) to determine the maximum and minimum?