Given $f\,(x) = \sqrt{25-x^2}$, show that the Mean Value Theorem applies to this function over the interval $[-3,5]$, and then do the following:
Determine if the Mean Value Theorem applies or does not apply to the given function on the given interval. When it applies, state clearly its conclusion, and then find the value(s) guaranteed to exist by the theorem. When it does not apply, state why.
$\displaystyle{f\,(x) = x^3 - x^2; \quad [-1,2]}$
$\displaystyle{f\,(x) = \frac{1}{x^3}; \quad [-1,1]}$
$\displaystyle{f\,(x) = |x+2|; \quad [-3,0]}$
$\displaystyle{f\,(x) = x^3-3x^2+2x; \quad [0,2]}$
$\displaystyle{f\,(x) = 8 - x^3; \quad [-2,1]}$
$\displaystyle{f\,(x) = 3x^{5/3} - 5x; \quad [-1,1]}$
$\displaystyle{f\,(x) = \frac{1}{x-1}; \quad [-2,1]}$
$\displaystyle{f\,(x) = x^2 - 5; \quad [-3,0]}$
$\displaystyle{f\,(x) = \sqrt{4-x^2}; \quad [0,2]}$
$\displaystyle{f\,(x) = 5x^{2/3} - 5; \quad [-1,1]}$
$\displaystyle{f\,(x) = x^{1/3} - 1; \quad [-1,1]}$
$\displaystyle{f\,(x) = x^3 - 1; \quad [-2,2]}$
$\displaystyle{f\,(x) = x^3 + 3x^2; \quad [-2,0]}$