Question

How do you find the global extreme values for `f(t) = tsqrt(4 - t²)` on [-1,2]?

Answer 1

Maximum is `(sqrt2,2)` and minimum is `(-1,-sqrt3)`.

Explanation:

The global extreme values of a function in a set `[a,b]` are to be searched between the local extreme values in the set `[a,b]` and in `a` or `b`.

The domain of the function is `D=[-2,2]`, because the radical argument has to be not-negative.

Let's search the local extreme:

`y'=1*sqrt(4-t^2)+t*1/(2sqrt(4-t^2))*(-2t)=`

`=sqrt(4-t^2)-t^2/sqrt(4-t^2)=(4-t^2-t^2)/sqrt(4-t^2)=2*(2-t^2)/sqrt(4-t^2)`.

Now:

`y'>=0`

if

`2-t^2>=0` (the radical is positive or zero in its domain!),

`-sqrt2<=t<=sqrt2`.

`f(-sqrt2)=-sqrt2*sqrt(4-2)=-2`

`f(sqrt2)=sqrt2*sqrt(4-2)=2`.

So the point `A(-sqrt2,-2)` is a local minimum and it is not in the set `[-1,2]` and the point `B(sqrt2,2)` is a local maximum that is in the set.

`f(-1)=-sqrt3`

`f(2)=0`.

SO:

The point `B(sqrt2,2)` is the global maximum and the point `C(-1,-sqrt3)` is the global minumum.

graph{xsqrt(4-x^2) [-10, 10, -5,5]}