Question

enlighten me:

If f(x) = (x^3)+x and h(x) is the inverse of f(x), then h'(2) is ??

Answer 1

The inverse of f(x) = h(x) = 1/((x^3)+x)
h'(x) = -3(x^-4) -(x^2)
h'(2) = -3(32) - (4)
h'(2) = -100

Answer 2

well...the five answer choices are as follows:
a) 1/13
b) 1/4
c) 1
d) 4
e) 13

Answer 3

I think what is meant here is function inverse h(f(x)) = x, not the multiplicative inverse that you meant.

Answer 4

the question was stated as it appears on the worksheet, its basically asking for the derivative of the inverse of the function f(x)

Answer 5

the answer is 1/6 i guess, follow my next post

Answer 6

i thought it was 1/13...if the answer choices mean anything at all then 1/6 is incorrect

Answer 7

this is what you do. the formula for finding the derivative of an inverse function is


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

Answer 8

so first find
\[f^{-1}(2)\] by solving
\[x^3+x=2\]
\[x=1\]

Answer 9

then find
\[f'(x)=3x^2+1\] then find
\[f'(1)=4\]
and so your answer is
\[(f^{-1}(2))'=\frac{1}{4}\]

Answer 10

hope steps are clear

Answer 11

ok, I couldn't find any better solution.

write \[f(x) = y, x = h(y);\]
then the equation becomes: \[y = h^3(y) + h(y)\]
h is differentiable (given), hence differentiate to get:
\[1 = (3 h^2(y) + 1 ) h'(y)\]
plug y = 2
\[1 = (3 h^2(2)+1) h'(2)\]
we also know that
\[h(f(x)) = x \Rightarrow h(x^3+x) = x \Rightarrow h(1+1) = 1\]
now put this in the previous expression and get
\[h'(2) = 1/(3\times1 + 1) = \frac{1}{4}\]

my last calculation was wrong, sorry