Find an equation for the line tangent to the graph of $f^{-1}$ at the point $(3,1)$ if $f(x)=x^3+2x^2-x+1$

Question

Find an equation for the line tangent to the graph of $f^{-1}$ at the point $(3,1)$ if $f(x)=x^3+2x^2-x+1$

ok, so I know that I need to take the derivative of f(x).

$f'(x)=3x^2+4x-1$

The inverse function is then supposed to be $\frac{1}{f'(f^{-1}(f(x)))}$ I think. I'm not sure if this is the right formula. Can someone help me verify if this is the correct formula. Otherwise I think I know how to wrap up the problem but I'm not sure if I'm using the correct formula and if so, how do I know what value to evaluate for x? Am i pluggin in 3 for x? Since they gave me the point (3,1) to use

Answer 1

You have\begin{align}(f^{-1})'(3)&=\frac1{f'\left(f^{-1}(3)\right)}\\&=\frac1{f'(1)}\\&=\frac16.\end{align}So, the tangent line is$$y=\frac16(x-3)+1=\frac x6+\frac12.$$

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Although this is not being asked, I think that it is interesting to note that the line $y=13-4x$ is also a tangent line to that curve passing through $(3,1)$; it is tangent at $\left(\frac{29}8,-\frac32\right)$ (see the picture below).

Answer 2

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

$$x=f(y)=y^3+2y^2-y+1$$

$$\dfrac{dx}{dy}=3y^2+4y-1$$

What is $$\dfrac{dy}{dx}$$ at $(3,1)?$

Answer 3

Differentiating $f(f^{-1}(x) ) = x $ gives $$ f’(f^{-1}(x)) \cdot (f^{-1})’ = 1 \implies (f^{-1})’ = \frac{1}{f’(f^{-1}(x))} $$ Now plug in $x=3$, because you need the slope of $f^{-1}$ at that point. It’s also given that $f^{-1}(3) =1$, so $$( f^{-1} )’(3,1) = \frac{1}{f’(1)} =\frac 16$$

Answer 4

If you were simply asked to find the equation of a tangent line for a given function $f(x)$ at some point $(x_0, y_0)$--in other words, no inverse functions involved--then it would be simple, right? You'd write $$y - y_0 = f'(x_0)(x - x_0).$$

But since the inverse function $f^{-1}$ is simply the reflection of $f$ about the line $y = x$, you just reflect the point of tangency by exchanging $x$ and $y$: $$(x_0, y_0) = (1,3)$$ and the tangent line to $f$ at this point is $$y - 3 = 6(x - 1),$$ since by your calculation, $$f'(x) = 3x^2 + 4x - 1$$ hence $$f'(x_0) = f'(1) = 3 + 4 - 1 = 6.$$ Now all that is left is to take the equation for the tangent line and interchange $x$ and $y$: $$x - 3 = 6(y - 1),$$ which in slope-intercept form is $$y = \frac{1}{6}(x+3).$$

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The idea here is that we perform the tangent line calculation for the original function, at a point corresponding to the desired point of tangency of the inverse function, by reflecting this point across the line $y = x$. Then once we have the tangent line for the original function, we reflect it back across $y = x$ to recover the tangent line for the inverse function at the originally desired point.