# 1.4 Inverse functions  (Page 2/11)

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## Rule: horizontal line test

A function $f$ is one-to-one if and only if every horizontal line intersects the graph of $f$ no more than once. (a) The function f ( x ) = x 2 is not one-to-one because it fails the horizontal line test. (b) The function f ( x ) = x 3 is one-to-one because it passes the horizontal line test.

## Determining whether a function is one-to-one

For each of the following functions, use the horizontal line test to determine whether it is one-to-one.

1. 2. 1. Since the horizontal line $y=n$ for any integer $n\ge 0$ intersects the graph more than once, this function is not one-to-one. 2. Since every horizontal line intersects the graph once (at most), this function is one-to-one. Is the function $f$ graphed in the following image one-to-one? No.

## Finding a function’s inverse

We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in the domain of $f$ to elements in the range of $f.$ The inverse function maps each element from the range of $f$ back to its corresponding element from the domain of $f.$ Therefore, to find the inverse function of a one-to-one function $f,$ given any $y$ in the range of $f,$ we need to determine which $x$ in the domain of $f$ satisfies $f\left(x\right)=y.$ Since $f$ is one-to-one, there is exactly one such value $x.$ We can find that value $x$ by solving the equation $f\left(x\right)=y$ for $x.$ Doing so, we are able to write $x$ as a function of $y$ where the domain of this function is the range of $f$ and the range of this new function is the domain of $f.$ Consequently, this function is the inverse of $f,$ and we write $x={f}^{-1}\left(y\right).$ Since we typically use the variable $x$ to denote the independent variable and $y$ to denote the dependent variable, we often interchange the roles of $x$ and $y,$ and write $y={f}^{-1}\left(x\right).$ Representing the inverse function in this way is also helpful later when we graph a function $f$ and its inverse ${f}^{-1}$ on the same axes.

## Problem-solving strategy: finding an inverse function

1. Solve the equation $y=f\left(x\right)$ for $x.$
2. Interchange the variables $x$ and $y$ and write $y={f}^{-1}\left(x\right).$

## Finding an inverse function

Find the inverse for the function $f\left(x\right)=3x-4.$ State the domain and range of the inverse function. Verify that ${f}^{-1}\left(f\left(x\right)\right)=x.$

Follow the steps outlined in the strategy.

Step 1. If $y=3x-4,$ then $3x=y+4$ and $x=\frac{1}{3}y+\frac{4}{3}.$

Step 2. Rewrite as $y=\frac{1}{3}x+\frac{4}{3}$ and let $y={f}^{-1}\left(x\right).$

Therefore, ${f}^{-1}\left(x\right)=\frac{1}{3}x+\frac{4}{3}.$

Since the domain of $f$ is $\left(\text{−}\infty ,\infty \right),$ the range of ${f}^{-1}$ is $\left(\text{−}\infty ,\infty \right).$ Since the range of $f$ is $\left(\text{−}\infty ,\infty \right),$ the domain of ${f}^{-1}$ is $\left(\text{−}\infty ,\infty \right).$

You can verify that ${f}^{-1}\left(f\left(x\right)\right)=x$ by writing

${f}^{-1}\left(f\left(x\right)\right)={f}^{-1}\left(3x-4\right)=\frac{1}{3}\left(3x-4\right)+\frac{4}{3}=x-\frac{4}{3}+\frac{4}{3}=x.$

Note that for ${f}^{-1}\left(x\right)$ to be the inverse of $f\left(x\right),$ both ${f}^{-1}\left(f\left(x\right)\right)=x$ and $f\left({f}^{-1}\left(x\right)\right)=x$ for all x in the domain of the inside function.

Find the inverse of the function $f\left(x\right)=3x\text{/}\left(x-2\right).$ State the domain and range of the inverse function.

${f}^{-1}\left(x\right)=\frac{2x}{x-3}.$ The domain of ${f}^{-1}$ is $\left\{x|x\ne 3\right\}.$ The range of ${f}^{-1}$ is $\left\{y|y\ne 2\right\}.$

## Graphing inverse functions

Let’s consider the relationship between the graph of a function $f$ and the graph of its inverse. Consider the graph of $f$ shown in [link] and a point $\left(a,b\right)$ on the graph. Since $b=f\left(a\right),$ then ${f}^{-1}\left(b\right)=a.$ Therefore, when we graph ${f}^{-1},$ the point $\left(b,a\right)$ is on the graph. As a result, the graph of ${f}^{-1}$ is a reflection of the graph of $f$ about the line $y=x.$

#### Questions & Answers

find the domain and range of f(x)= 4x-7/x²-6x+8
Nick Reply
find the range of f(x)=(x+1)(x+4)
Jane Reply
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
Harrieta Reply
d/dx{1/y - lny + X^3.Y^5}
mogomotsi Reply
How to identify domain and range
Umar Reply
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
Adri
sorry
Dr
hi adri ana
Dr
:(
Shun
was up
Dr
hello
Adarsh
is it chatting app?.. I do not see any calculus here. lol
Adarsh
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie

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