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

Question

4770 views around the world

You can reuse this answer
Creative Commons License

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]}