Answer
-$\frac{1}{6}$
Work Step by Step
$f(x)$ = $5-2x^{3}$, $a$ = $7$
$5-2x^{3}$ = $7$
$2x^{3}+2$ = $0$
$2(x+1)(x^{2}-x+1)$ = $0$
$x$ = -$1$
$f(-1)$ = $7$ $so$ $f^{-1}$$(7)$ = -$1$
($f^{-1}$)'$(7)$ = $\frac{1}{f'(f^{-1}(7))}$ = $\frac{1}{f'(-1)}$
$f'(x)$ = -$6x^{2}$
$f'(-1)$ = -$6(-1)^{2}$ = -$6$
$so$ ($f^{-1}$)'$(7)$ = $\frac{1}{f'(-1)}$ = -$\frac{1}{6}$
