I have a question that states if $f(x) = x^3+3x-1$ from $(-\infty,\infty)$ calculate $g'(3)$using the formula $$ g'(x)= \left(\frac1{f'(g(x))}\right )$$
If I am thinking about this correctly does that mean that the slope of $f(x)$ is the reciprocal of its inverse function $g(x)$
If so then can I just differentiate $f(x)$, plug in 3, and find the reciprocal of that to find $g'(3)$?
$$f'(x) = 3x^2 + 3$$
$$f'(3) = 30$$
$$g'(3) = \frac1{30}$$
Best Answer
Not quite. First you gave the correct formula $$ g'(x) = \frac{1}{f'(g(x))} \quad (*) $$ which by the way is a consequence of $f(g(x)) = x$ and taking the derivative of both sides.
You then applied the wrong formula $$ g'(3) = \frac{1}{f'(3)} \mbox{vs.} \frac{1}{f'(g(3))} $$ which will not work in general except if $g(3) = 3$ which is not the case here:
$$ f(x) = x^3 + 3x - 1 = 3 $$ has the only real solution $x = 1$. So we have $f(1) = 3$ which means $g(3) = 1$.
This leads to $$ g'(3) = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2+3} = \frac{1}{6} $$
Compare this with the graphs:
(Larger version of the image)