Answer
$\frac{1}{17}$
Work Step by Step
$f(x)$ = $\frac{1}{27}$($x^{5}+2x^{3})$, $a$ = -$11$
$\frac{1}{27}$($x^{5}+2x^{3})$ = -$11$
$x^{5}+2x^{3}+11$ = $0$
$(x+3)(x^{4}-3x^{3}+11x^{2}-33x+99)$ = $0$
$x$ = -$3$
$f(-3)$ = -$11$ $so$ $f^{-1}(-11)$ = -$3$
$(f^{-1})$'$(-11)$ = $\frac{1}{f'(f^{-1}(-11))}$ = $\frac{1}{f'(-3)}$
$f'(x)$ = $\frac{5x^{4}}{27}$ + $\frac{6x^{2}}{27}$
$f'(-3)$ = $\frac{5(-3)^{4}}{27}$ + $\frac{6(-3)^{2}}{27}$ = $15$ + $2$ = $17$
$so$ $(f^{-1})$'$(-11)$ = $\frac{1}{f'(-3)}$ = $\frac{1}{17}$
