Question

Find (f^-1)'(a) for
Sqrt(x^3+x^2+x+1), a=2

Answer 1

\[find (f^-1)'(a) for \sqrt{x ^{3}+ x^2+x+1} \]

Answer 2

where a=2

Answer 3

my guess is \(f^{-1}(2)=1\)

Answer 4

what inverse equation did you get for that?

Answer 5

i set
\[\sqrt{x^3+x^2+x+1}=2\] and made an educated guess as to what \(x\) should be

Answer 6

since \(f(1)=\sqrt{4}=2\) that tells you \(f^{-1}(2)=1\)

Answer 7

Makes sense

Answer 8

oh i wasn't looking carefully

Answer 9

But i feel theres a more effective way to prove this

Answer 10

you don't want \(f^{-1}(2)\) you want the derivative at \(2\)

Answer 11

the derivative of the inverse, correct?

Answer 12

no matter,
\[f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}\]

Answer 13

in this case
\[f^{-1}(2)=\frac{1}{f;(f^{-1}(2))}=\frac{1}{f'(1)}\]

Answer 14

so your job is
1) find the derivative of \(f(x)=\sqrt{x^3+x^2+x+1}\)
2) find \(f'(1)\)
3) take the reciprocal

Answer 15

I suppose my confusion rests in how the inverse function of f=1, or did that come from setting it equal to A?

Answer 16

you can almost do it in your head
\[f(x)=\sqrt{x^3+x^2+x+1}\]
\[f'(x)=\frac{3x^2+2x+1}{2\sqrt{x^3+x^2+x+1}}\]
\[f'(1)=\frac{6}{4}=\frac{3}{2}\] and the reciprocal of \(\frac{3}{2}\) is \(\frac{2}{3}\)

Answer 17

how did if find \(f^{-1}(2)\)?
i think it is called "by inspection"
don't try using algebra
just think

Answer 18

ah, all right- i had the order of operations mixed up it seems