# Find the slope of the tangent line to the function's inverse

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## Homework Statement

For ##y=f(x)##,
find the slope of the tangent line to its inverse function ##f^{-1}## at the indicated point P.

##f(x) = -x^3-x+2## , ##P(-8,2)##

## Homework Equations

The Inverse Function Theorem:
##(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}##

## The Attempt at a Solution

So I'm stuck at square one. I need to find the inverse of ##f## and am having troubles doing so.
The given is ##f(x) = -x^3-x+2## and I'm at a stand-still in isolating the x.

##f(x) = -x^3-x+2##
##y = -x^3-x+2##
##y-2 = -x^3-x##
##y-2 = -x(x^2+1)##
.
.
?

Could someone poke my brain a little on this one?

Orodruin
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I need to find the inverse of fff and am having troubles doing so.
No you don’t. That is the point of the exercise. Use the relevant equation you posted.

opus
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So it's to my understanding that between ##f## and ##f'##, their derivatives should be reciprocals to each other according to the posted equation.
Following your advice, I used to formula and took the derivative of ##f## and got ##f'(x)=-3x^2-1##. Next I plugged in the x value to the point, x=-8, into ##f'## and got -193. So this would tell me that the derivative of its inverse would be ##\frac{-1}{193}##. But in the back of the book it says ##\frac{-1}{96}## so I'm sure I have misunderstood something about this problem.

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Work so far:

(i) I want the slope of ##f^{-1}##. I dont know what ##f^{-1}## is, but that doesn't matter because as long as I know what the slope of ##f## is, I can find the slope of ##f^{-1}##.

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

(iii) By the Inverse Function Theorem, ##(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}##
I know what ##f'## is.
Since ##f## and ##f^{-1}## are inverses, their x and y values are swapped.
Point ##P = (-8,2)## which corresponds to ##f##, so when the Inverse Function Theorem is asking me for ##f^{-1}(x)##, it wants the -8.

So,
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{-3(-8)^2-1} = \frac{-1}{193}$$

So the slope of the tangent line to ##f^{-1}## at point P should be ##\frac{-1}{193}##
However, the solution key says that it should be ##\frac{-1}{96}##

Orodruin
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The point ##x = -8## and ##y = 2## does not satisfy ##f(x) = y##. However, it does satisfy ##x = -8##, ##y = f^{-1}(x) = 2## so what you need is ##1/f'(2)##. The result is neither -1/193 nor -1/96. Are you looking at the correct place in the key?

opus
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The point x=−8x = -8 and y=2y = 2 does not satisfy f(x)=yf(x) = y. However, it does satisfy x=−8x = -8, y=f−1(x)=2
What do you mean by this? Are you saying that the given point ##P(-8,2)## goes with ##f^{-1}## and not ##f##, and that the corresponding point to ##f## would be ##Q(2,-8)##?

The result is neither -1/193 nor -1/96. Are you looking at the correct place in the key?
Yes I'm certain. After not being able to come up with ##\frac{-1}{96}##, I moved on to part B of the question.
Part B states: Find the equation of the tangent line to the graph of ##f^{-1}## at the indicated point.
The solution in the back of the book to this problem is ##y=\frac{-1}{13}x+\frac{18}{3}## Having the slope of this line be ##\frac{-1}{13}## makes me think that there's yet another typo in this book.
I went back and evaluated ##f'(2)## into part A, and this gave me ##\frac{-1}{13}##. So I think this shows that my statement about evaluating at 8 was incorrect. I guess I'm unclear on whether point ##P## is on the graph of ##f## or ##f^{-1}##. I know they're the same graph, just reflected when we switch the x and y coordinates, but in doing so, the domains and ranges swap.

Orodruin
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It is an easy check to see if (-8,2) and (2,-8) are on the graphs of ##f## or that of ##f^{-1}## (just insert the values and see if they satisfy ##(x,f(x))## or ##(f(y),y)##). In this case, (-8,2) is on the graph of ##f^{-1}## as ##(f(2),2) = (-2^3-2+2,2)=(-8,2)##.

Plotting the result of part B is a good way of double checking your result (although I would put it on the form ##y = k (x-x_0)+y_0## - in this case with ##x_0 = -8## and ##y_0 = 2##). You would obtain this for the answer key to B:

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opus
Ray Vickson
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Work so far:

(i) I want the slope of ##f^{-1}##. I dont know what ##f^{-1}## is, but that doesn't matter because as long as I know what the slope of ##f## is, I can find the slope of ##f^{-1}##.

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

(iii) By the Inverse Function Theorem, ##(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}##
I know what ##f'## is.
Since ##f## and ##f^{-1}## are inverses, their x and y values are swapped.
Point ##P = (-8,2)## which corresponds to ##f##, so when the Inverse Function Theorem is asking me for ##f^{-1}(x)##, it wants the -8.

So,
$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{-3(-8)^2-1} = \frac{-1}{193}$$

So the slope of the tangent line to ##f^{-1}## at point P should be ##\frac{-1}{193}##
However, the solution key says that it should be ##\frac{-1}{96}##

The tangent line to the graph of ##y = f(x)## at ##x=2, y=-8## has equation ##y+8 = -13(x-2),## since ##f'(2) =-13.## So, the tangent line to the inverse function ##x = g(y)## is obtained by solving for ##x## in terms of ##y## in the original tangent line. That will give you ##g'(y)= (f^{-1})'(y)## at ##y = -8##.

I have never liked the practice of using ##x## in both ##f(x)## and ##f^{-1}(x)##; to me, it is much clearer (and less confusing) to use ##f(x)## and ##f^{-1}(y).## That is also consistent with what we do in fields such as topology.

Last edited:
opus
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It is an easy check to see if (-8,2) and (2,-8) are on the graphs of ff or that of f−1
Well this was part of my original troubles. I couldn't find the actual inverse so I couldn't plot the inverse graph. I didn't know how to fully isolate the x term.
I plotted ##f## and (2,-8) was on it, but not (8,-2). In imagining it's reflection, (since I didn't have it's function), (8,-2) was in fact on the reflection. So now it's clear to me that the given point ##P(-8,2)## belongs to ##f^{-1}## not ##f##.
What website did you use to plot that graph so you could see the derivative? I use Desmos and I don't think it has that option.

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The tangent line to the graph of ##y = f(x)## at ##x=2, y=-8## has equation ##y+8 = -13(x-2),## since ##f'(2) =-13.## So, the tangent line to the inverse function ##x = g(y)## is obtained by solving for ##x## in terms of ##y## in the original tangent line. That will give you ##g'(y)= (f^{-1})'(y)## at ##y = -8##.

I have never liked the practice of using ##x## in both ##f(x)## and ##f^{-1}(x)##; to me, it is much clearer (and less confusing) to use ##f(x)## and ##f^{-1}(y)##. That is also consistent with what we do in fields such as topology.
Inverses always twist my brain in knots. I don't know what x goes to what, the given a's, y's etc etc. Every time I look at a value I have to trace it's origins all the way back to the question to figure out where it belongs. It's really confusing notation.

Ray Vickson
Homework Helper
Dearly Missed
Well this was part of my original troubles. I couldn't find the actual inverse so I couldn't plot the inverse graph. I didn't know how to fully isolate the x term.
I plotted ##f## and (2,-8) was on it, but not (8,-2). In imagining it's reflection, (since I didn't have it's function), (8,-2) was in fact on the reflection. So now it's clear to me that the given point ##P(-8,2)## belongs to ##f^{-1}## not ##f##.
What website did you use to plot that graph so you could see the derivative? I use Desmos and I don't think it has that option.

Easiest way: plot the graph of ##y = f(x)## on an actual, physical sheet of graph paper, then turn it through 90°. That will give you the plot of the inverse function.

opus
Gold Member
Easiest way: plot the graph of ##y = f(x)## on an actual, physical sheet of graph paper, then turn it through 90°. That will give you the plot of the inverse function.
Hey that's a great idea. I'm gonna use that. Thanks!

Orodruin
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Well this was part of my original troubles. I couldn't find the actual inverse so I couldn't plot the inverse graph.
You do not need the functional form of the inverse to plot it. As you said previously, just switch the axes. How to do this depends on the program you are using to plot.

What website did you use to plot that graph so you could see the derivative?
I used Matlab.

Ray Vickson