Find the value of $\frac{df^{-1}}{dx}$ of the following function
Suppose that the differentiable function $y = f(x)$ has an inverse and that the graph of ƒ passes through the point $(2, 4)$ and has a slope of $\frac13$ there. Find the value of $\frac{df^{-1}}{dx}$ at $x = 4 $
The answer in the textbook says it's $3$ but I keep getting different answers everytime I attempt the question.
My solution:
Using $(2,4)$ and $m = \frac13$ I got $y = \frac13x + \frac{10}3$
Plugging $x = 4$ into the equation, I get $(4, \frac{14}3)$
Using $(f^{-1})'(b) = \frac{1}{f'(a)}$ I get $\frac19$, which is obviously different from the textbook's answer.
Can someone help push me in the right direction?
$\endgroup$2 Answers
$\begingroup$Since $y=f(x)$ passes through the point $(2,4)$, we can say that $f(2)=4$.
Therefore, $f^{-1}(4)=2$. Now$$\dfrac{df^{-1}}{dx}\bigg|_{x=4}=\dfrac{1}{\dfrac{df}{dx}\bigg|_{x=f^{-1}(4)=2}}=\dfrac{1}{\dfrac13}=3$$
So, the value of $\dfrac{df^{-1}}{dx}$ at $x=4$ is $3$.
$\endgroup$ $\begingroup$You don't need the equation of the tangent line. One simply applies the formula that$$ (f^{-1})'(f(a))=\frac{1}{f'(a)} $$where in the problem $a=2$, $f(a)=4$ and $f'(a)=\frac13$.
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