17Calculus Precalculus - Inverse Functions
This page covers inverse functions in general. Inverses of trig functions are discussed on the trig inverse page.
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The idea of inverse functions is that if you input a value into one function, then take that result and input it into the inverse function, you should get the original value back. You could also think of the inverse function undoing the original function. Figure 1 should help you visualize what is going on.
Figure 1 |
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The idea is that \(x\) is input to the function \(f(x)\) with output y. This \(y\) is then the input of the inverse function, whose output is the orginal \(x\) that we started with. Pretty cool, eh? A couple of comments are in order.
1. We use the notation \(f^{-1}(y)\) to describe the inverse of \(f(x)\). In this context, \(f^{-1}(y) \neq 1/f(y)\). You can usually tell by the context what is meant when you run across this notation.
2. This loop works in both directions, i.e. if you reverse the direction of the arrows and input x into the inverse function first, the result is the same.
Here is a great video explaining inverse functions and where we are going with them.
PatrickJMT - Inverse Functions - The Basics [16min-59secs]
video by PatrickJMT |
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Horizontal Line Test
A graph that passes the horizontal line test and, therefore, has an inverse is called one-to-one.
You are probably familiar with the vertical line test that tells you whether the graph is a function or not. Well, a similiar rule holds called the horizontal line test. When you have a graph and you draw horizontal lines, if all horizontal lines pass through the graph only once, then the graph is invertible, i.e. an inverse exists. If you can draw at least one line that passes through the graph more than once, then the graph is NOT invertible and no inverse exists.
One thing to note is that this tells you whether or an inverse exists but not how to find it. In fact, you may not be able to find the inverse even if the horizontal line test tells you that one exists.
Here is a good video with lots of examples, explaining the horizontal line test. He also mentions the vertical line test, which helps you see the parallel.
MIP4U - Determine if the Graph of a Relation is a One-to-One Function [3min-36secs]
video by MIP4U |
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When a graph does not pass the horizontal line test, we can often restrict the domain, so that the graph does pass the horizontal line test and then have an inverse. (This is especially helpful in the case of trigonometric functions.) For example, you know that the parabola \(y=x^2\) does not pass the horizontal line test. However, we can restrict the domain by requiring that \(x > 0\). We now have a graph that passes the horizontal line test and therefore has an inverse.
Finding An Inverse
Finding an inverse of a function is pretty straightforward. There are some specific steps to follow but the idea is that you change all x's to y's and all y's to x's and then solve the new equation for y. Rather than going into a lot of detail, this video explains it very well.
This video goes into a lot of detail with inverse functions and gives the entire picture. He also has some really good examples.
MIP4U - Inverse Functions [9min-40secs]
video by MIP4U |
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Something you always want to do in math is try to check your answer. The way you check your answer after calculating the inverse is to form two composite functions and see what you get. For example, if your original function was \(f(x)\), we can write the inverse as \(f^{-1}(x)\). The notation here can get confusing. Since the context is inverse functions, \(f^{-1}(x) \neq 1/f(x)\).
The two composite functions are \(f(f^{-1}(x))\) and \(f^{-1}(f(x))\). In both cases, the result should be x, if they are truly inverses of each other. Note that you need to calculate BOTH composite functions, since, if you made a mistake, it is possible for one to evaluate to x and the other not be equal to x.
If you need to review composite functions, you can find more explanation and practice problems on the composite functions page.
Graph Of An Inverse Function
Figure 2 |
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The graph of an inverse function is interesting when compared to the orginal function. The graphs are reflected about the line \(y=x\). If you think about it, it makes sense. After all, in the discussion above, we switch the x's and y's as one of the steps when finding an inverse. Figure 2 shows an example. The two curved graphs are inverses of each other. The black straight line is the line \(y=x\). Notice the nice symmetry across the line.
Okay, time for some practice problems.
Practice
Instructions - - Unless otherwise instructed, find the inverse of these functions without using a calculator and give your answers in exact, simplified form.
Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{x+4}-3\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{x+4}-3\) without using a calculator and give your answer in exact, simplified form.
Solution |
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1559 video
video by PatrickJMT |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{y=\frac{5x-3}{2x+1}}\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{y=\frac{5x-3}{2x+1}}\) without using a calculator and give your answer in exact, simplified form.
Solution |
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1560 video
video by PatrickJMT |
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Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{x-2}\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{x-2}\) without using a calculator and give your answer in exact, simplified form.
Solution |
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1566 video
video by Krista King Math |
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Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{-1-x}\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(f(x)=\sqrt{-1-x}\) without using a calculator and give your answer in exact, simplified form.
Solution |
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1567 video
video by Krista King Math |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x - 7 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x - 7 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = (x + 7)/2 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x - 7 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2903 video
Final Answer |
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\( f^{-1}(x) = (x + 7)/2 \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = x^3 + 8 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = x^3 + 8 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = \sqrt[3]{x-8} \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = x^3 + 8 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2904 video
Final Answer |
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\( f^{-1}(x) = \sqrt[3]{x-8} \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{x+2} - 5 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{x+2} - 5 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = x^2 + 10x + 23 \) The instructor does not say anything about the domain of the inverse function. He needs to specify that for \( f^{-1}(x) \) the domain is \( x \geq -5 \). This is important since your instructor may take off points if you do not specify the domain here.
How do you get this domain? Well, we got from the graph but if you don't have the graph, you know the domain of \(f(x)\) is \(x \geq -2 \) since you cannot take the square root of negative numbers. Now, the range of \(f(x) \) is \( y \geq -5 \). In this case, the range of \(f(x) \) becomes the domain of \(f^{-1}(x)\). So the domain of the invers function is \( x \geq -5 \).
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{x+2} - 5 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2905 video
Final Answer |
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\( f^{-1}(x) = x^2 + 10x + 23 \) The instructor does not say anything about the domain of the inverse function. He needs to specify that for \( f^{-1}(x) \) the domain is \( x \geq -5 \). This is important since your instructor may take off points if you do not specify the domain here. |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x+4} - 2 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x+4} - 2 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = (x+2)^3 - 4 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x+4} - 2 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2906 video
Final Answer |
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\( f^{-1}(x) = (x+2)^3 - 4 \) |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{3x-7}{4x+3} }\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{3x-7}{4x+3} }\) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{3x+7}{3-4x} }\) The instructor does not specify the domain of this inverse function. We need to restrict the domain to all real numbers except \( x = 3/4 \) and \( x = -3/4 \). The first number occurs when the denominator of the inverse function is zero. The second number occurs when the denominator of \( f(x) \) is zero. Both values need to specified. However, check with your instructor to see what they require.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{3x-7}{4x+3} }\) without using a calculator and give your answer in exact, simplified form.
Solution |
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2907 video
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{3x+7}{3-4x} }\) The instructor does not specify the domain of this inverse function. We need to restrict the domain to all real numbers except \( x = 3/4 \) and \( x = -3/4 \). The first number occurs when the denominator of the inverse function is zero. The second number occurs when the denominator of \( f(x) \) is zero. Both values need to specified. However, check with your instructor to see what they require. |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{x+2}{x} }\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{x+2}{x} }\) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{2}{x-1} }\) with domain \( \{ x | x \in \mathbb{R}, x \neq 0, x \neq 1 \} \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{x+2}{x} }\) without using a calculator and give your answer in exact, simplified form.
Solution |
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2908 video
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{2}{x-1} }\) with domain \( \{ x | x \in \mathbb{R}, x \neq 0, x \neq 1 \} \) |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{2x+5}{3x-1} }\) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{2x+5}{3x-1} }\) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{3x+5}{3x-2} }\) with domain \( \{ x | x \in \mathbb{R}, x \neq 1/3, x \neq 2/3 \} \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \(\displaystyle{ f(x) = \frac{2x+5}{3x-1} }\) without using a calculator and give your answer in exact, simplified form.
Solution |
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2909 video
Final Answer |
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\(\displaystyle{ f^{-1}(x) = \frac{3x+5}{3x-2} }\) with domain \( \{ x | x \in \mathbb{R}, x \neq 1/3, x \neq 2/3 \} \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{2x-6} \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{2x-6} \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = x^2 / 2 + 3 \) for \( x > 0 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt{2x-6} \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2910 video
Final Answer |
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\( f^{-1}(x) = x^2 / 2 + 3 \) for \( x > 0 \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x-4} + 1 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x-4} + 1 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = (x-1)^3 + 4 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = \sqrt[3]{x-4} + 1 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2911 video
Final Answer |
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\( f^{-1}(x) = (x-1)^3 + 4 \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = e^{3x+1} - 5 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = e^{3x+1} - 5 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = (1/3)\ln(x-5) - 1/3 \) with domain \( x > 5 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = e^{3x+1} - 5 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2912 video
Final Answer |
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\( f^{-1}(x) = (1/3)\ln(x-5) - 1/3 \) with domain \( x > 5 \) |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x-3 \) without using a calculator and give your answer in exact, simplified form.
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x-3 \) without using a calculator and give your answer in exact, simplified form.
Final Answer |
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\( f^{-1}(x) = (x+3)/2 \)
Problem Statement |
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Unless otherwise instructed, find the inverse of the function \( f(x) = 2x-3 \) without using a calculator and give your answer in exact, simplified form.
Solution |
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2913 video
Final Answer |
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\( f^{-1}(x) = (x+3)/2 \) |
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inverse functions 17calculus youtube playlist
Here is a playlist of the videos on this page.
Really UNDERSTAND Precalculus
Topics You Need To Understand For This Page
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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