What is the absolute maximum of y= sqr root 9-x^2 on the interval [-6,6]? Thanks

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Answer 1 · 2018-03-11T18:02:38

Please see below.

Because #sqrt(9-x^2)# has domain #[-3,3]#, I assume the function is

#f(x) = sqrt9 - x^2 = 3-x^2# which is a parabola that opens downward and has vertex on the #y# axis.

So, the maximum is #3# at #x=0#.

graph{3-x^2 [-11.25, 11.25, -5.625, 5.625]}

Perhaps you are working with a definition of maximum that allows us to ask for a maximum on some set other that the domain of a function.

#y= sqrt(9-x^2)# is the upper half of the circle whose equation is #x^2+y^2=9#.

With center #(0,0)# and radius #3#, the maximum #y# value is #3#.

graph{sqrt(9-x^2) [-11.25, 11.25, -5.625, 5.625]}