will include all the local maxima, minima, and some of the points of inflection.
Evaluate f(t) at these points, then the global maxima, will be the largest of
these and the value of f(t) at the end points of [-2,4], similarly for the
The derivative of f is:
f'(t)= 2 sqrt(16-t^2)-2t^2/sqrt(16-t^2)
and f'(t)=0 has roots at +/-2 sqrt(2), the negative root is outside the
interval of interest, and f(2\sqrt(2))=16. (note possible exceptional points
at t=+/-4 which may need to be handled seperatly)
Now at the end points of the inteval we have f(-2)=-8 sqrt(3), and f(4)=0.
Hence the global maximum on [-2,4] is at t=2sqrt(2), and is 16, and the global
minimum is at t=-2 and is -8sqrt(3).
The attachment shows a plot of the function over the specified interval