3.7 Inverse functions  (Page 6/9)

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Draw graphs of the functions and from [link] .

Is there any function that is equal to its own inverse?

Yes. If $\text{\hspace{0.17em}}f={f}^{-1},\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}f\left(f\left(x\right)\right)=x,\text{\hspace{0.17em}}$ and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because

$\frac{1}{\frac{1}{x}}=x$

Any function $\text{\hspace{0.17em}}f\left(x\right)=c-x,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$ is a constant, is also equal to its own inverse.

Access these online resources for additional instruction and practice with inverse functions.

Visit this website for additional practice questions from Learningpod.

Key concepts

• If $\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$ is the inverse of $\text{\hspace{0.17em}}f\left(x\right),\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)=x.\text{\hspace{0.17em}}$ See [link] , [link] , and [link] .
• Only some of the toolkit functions have an inverse. See [link] .
• For a function to have an inverse, it must be one-to-one (pass the horizontal line test).
• A function that is not one-to-one over its entire domain may be one-to-one on part of its domain.
• For a tabular function, exchange the input and output rows to obtain the inverse. See [link] .
• The inverse of a function can be determined at specific points on its graph. See [link] .
• To find the inverse of a formula, solve the equation $\text{\hspace{0.17em}}y=f\left(x\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as a function of $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Then exchange the labels $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ See [link] , [link] , and [link] .
• The graph of an inverse function is the reflection of the graph of the original function across the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ See [link] .

Verbal

Describe why the horizontal line test is an effective way to determine whether a function is one-to-one?

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that $\text{\hspace{0.17em}}y$ -values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no $\text{\hspace{0.17em}}y$ -values repeat and the function is one-to-one.

Why do we restrict the domain of the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ to find the function’s inverse?

Can a function be its own inverse? Explain.

Yes. For example, $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{x}\text{\hspace{0.17em}}$ is its own inverse.

Are one-to-one functions either always increasing or always decreasing? Why or why not?

How do you find the inverse of a function algebraically?

Given a function $\text{\hspace{0.17em}}y=f\left(x\right),\text{\hspace{0.17em}}$ solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Interchange the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Solve the new equation for $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ The expression for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the inverse, $\text{\hspace{0.17em}}y={f}^{-1}\left(x\right).$

Algebraic

Show that the function $\text{\hspace{0.17em}}f\left(x\right)=a-x\text{\hspace{0.17em}}$ is its own inverse for all real numbers $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$

For the following exercises, find $\text{\hspace{0.17em}}{f}^{-1}\left(x\right)\text{\hspace{0.17em}}$ for each function.

$f\left(x\right)=x+3$

${f}^{-1}\left(x\right)=x-3$

$f\left(x\right)=x+5$

$f\left(x\right)=2-x$

${f}^{-1}\left(x\right)=2-x$

$f\left(x\right)=3-x$

$f\left(x\right)=\frac{x}{x+2}$

${f}^{-1}\left(x\right)=\frac{-2x}{x-1}$

$f\left(x\right)=\frac{2x+3}{5x+4}$

For the following exercises, find a domain on which each function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ restricted to that domain.

$f\left(x\right)={\left(x+7\right)}^{2}$

domain of $f\left(x\right):\text{\hspace{0.17em}}\left[-7,\infty \right);\text{\hspace{0.17em}}{f}^{-1}\left(x\right)=\sqrt{x}-7$

$f\left(x\right)={\left(x-6\right)}^{2}$

$f\left(x\right)={x}^{2}-5$

domain of $\text{\hspace{0.17em}}f\left(x\right):\text{\hspace{0.17em}}\left[0,\infty \right);\text{\hspace{0.17em}}{f}^{-1}\left(x\right)=\sqrt{x+5}$

Given $\text{\hspace{0.17em}}f\left(x\right)={x}^{3}-5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)=\frac{2x}{1-x}:$

1. Find $\text{\hspace{0.17em}}f\left(g\left(x\right)\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(f\left(x\right)\right).$
2. What does the answer tell us about the relationship between $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)?$

a. and $\text{\hspace{0.17em}}g\left(f\left(x\right)\right)=x.\text{\hspace{0.17em}}$ b. This tells us that $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
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Abhi
I rally confuse this number And equations too I need exactly help
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salma
Commplementary angles
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Sherica
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Tamia
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Uday
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salma
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opoku
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Ali
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Nharnhar