Given $f(2)=1$, $f’(2)=3$, and $f’’(2)=e$, find the first and second derivative of $f^{-1}(x)$ at $x=1$.

Question

I was studying calculus and bumped into this exercise:

Let $f(x)$ be a differentiable, one-to-one function. Suppose $f(2)=1$, $f’(2)=3$, and $f’’(2)=e$, find the first and second derivative of $f^{-1}(x)$ at $x=1$.

Here is my thought:

I think I should find out what $f(x)$ is, and compute it’s inverse function subsequently. However I couldn’t figure out what $f(x)$ could be. Also, I noticed that $f^{-1}(1)=2$, which may help me find the answer. Any suggestions? Or am I overlooking something important? Thanks a lot.

Answer 1

Hint:

Let $g(x)$ be the inverse function of $f(x)$. Note that $$g(f(x))=x$$ and apply Chain Rule twice to find $g'(1)$ and $g''(1)$ respectively.

Note: It's sometimes pretty hard to find out $f(x)$ when you are given $f(a)$, $f'(a)$ and $f''(a)$ instead of $f(a)$, $f(b)$ and $f(c)$. Therefore, it should be better to turn to the original meaning of an inverse function.

Answer 2

Here is an idea, we know that by definition of inverse,

$$ f(f^{-1} (x) ) = x $$

Suppose we differentiated both sides using the chain rule on left,

$$ \frac{df(x)}{dx}|_{f^{-1} (x) } \frac{df^{-1} (x) }{dx} = 1 \tag{1}$$

Now, what would happen if you plugged $x=1$ into the equation above?

For the second derivative, let's differentiate our previously derived identity(1) once again with the product rule,

$$ \frac{d^2 f(x) }{dx^2}|_{f^{-1} (x) } ( \frac{df^{-1} (x) }{dx})^2 + \frac{df}{dx}|_{f^{-1}(x) } \frac{d^2 f^{-1} (x) }{dx^2} = 0 \tag{2}$$

And, we have related the inverse function's derivative with the function's derivatives

Hope this helps!

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Let's do better! If we rearrange (1),

$$ \frac{df^{-1} (x) }{dx} = \frac{1}{ \frac{df}{dx}|_{f^{-1} (x) } } $$

And we can plug the above identity into the second factor of first term in $(2)$ which leads us to:

$$ \frac{d^2 f(x) }{dx^2}|_{f^{-1} (x) } ( \frac{1}{ \frac{df}{dx}|_{f^{-1} (x) } } )^2 + \frac{df}{dx}|_{f^{-1}(x) } \frac{d^2 f^{-1} (x) }{dx^2} = 0 \tag{3}$$

Now we have expressed the inverse function's derivative completely in the derivatives of the function.

Answer 3

hint

Let $$g(x)=f^{-1}(x)$$

we know that $$g'(x)=\frac{1}{f'(g(x))}$$

thus $$g'(1)=\frac{1}{f'(g(1))}$$

but

$$g(1)=2$$ and $$f'(2)=3$$ then

$$g'(1)=\frac{1}{f'(2)}=\frac 13$$

Differentiating the formula above, we get

$$g''(x)=-f''(g(x))g'(x).(g'(x))^2$$ So, $$g''(1)=-f''(g(1))(g'(1))^3$$ $$=-\frac{1}{27}f''(2)=-\frac{e}{27}$$

Answer 4

Let $y=f^{-1}(x)$. Then $f(y)=x$ so differentiating with respect to $x$, we find that $\frac{d}{dx}=\frac{d}{dy}\frac{dy}{dx}$. Hence, $f'(y)y'=1$ so $y'=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$. From this result, it should be obvious how to get $(f^{-1}(x))'$ at $x=1$.

Differentiate again. \begin{align} y''&=-(f'(f^{-1}(x)))^{-2}[f'(f^{-1}(x))]'\\ &=-(f'(f^{-1}(x)))^{-2}f''(f^{-1}(x))y'. \end{align} Hence, you know everything. Just plug in.