Question

Let f (x) = 9x + sqrt(x + 82).
find (f^-1)'(0)

Answer 1

\[(f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}\]

Answer 2

\[f(x)=y \rightarrow f^{-1}(y)=x \\ \text{ so letting } y=0 \text{ we need \to find } x \\ \text{ for when we have } \\ f(x)=0 \\ 9x+\sqrt{x+82}=0\]

Answer 3

for what x is this true

Answer 4

solve for x and let me know what you get

Answer 5

x=-1

Answer 6

ok so you have \[f(-1)=0 \rightarrow f^{-1}(0)=-1 \\ \text{ so now you have } (f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}=\frac{1}{f'(-1)}\]

Answer 7

now you just need to differentiate f(x) and
then plug in -1

Answer 8

so 1/(9+1/2(81)^-1/2)

Answer 9

\[f(x)=9x+\sqrt{x+82} \\ f'(x)=9+\frac{1}{2 \sqrt{x+82}} \\ f'(-1)=9+\frac{1}{2 \sqrt{-1+82}} =9+\frac{1}{2 \sqrt{81}}=9+\frac{1}{2 (9)}=9+\frac{1}{18}\]