# 1.7 Inverse functions  (Page 3/10)

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When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x}\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}{f}^{-1}\left(x\right)={x}^{2},\text{\hspace{0.17em}}$ because a square “undoes” a square root; but the square is only the inverse of the square root on the domain $\text{\hspace{0.17em}}\left[0,\infty \right),\text{\hspace{0.17em}}$ since that is the range of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{x}.$

We can look at this problem from the other side, starting with the square (toolkit quadratic) function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}.\text{\hspace{0.17em}}$ If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}\text{\hspace{0.17em}}$ with its range limited to $\text{\hspace{0.17em}}\left[0,\infty \right),\text{\hspace{0.17em}}$ which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If $\text{\hspace{0.17em}}f\left(x\right)={\left(x-1\right)}^{2}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[1,\infty \right),\text{\hspace{0.17em}}$ then the inverse function is $\text{\hspace{0.17em}}{f}^{-1}\left(x\right)=\sqrt{x}+1.$

• The domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ = range of $\text{\hspace{0.17em}}{f}^{-1}\text{\hspace{0.17em}}$ = $\text{\hspace{0.17em}}\left[1,\infty \right).$
• The domain of $\text{\hspace{0.17em}}{f}^{-1}\text{\hspace{0.17em}}$ = range of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ = $\text{\hspace{0.17em}}\left[0,\infty \right).$

Is it possible for a function to have more than one inverse?

No. If two supposedly different functions, say, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ both meet the definition of being inverses of another function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ then you can prove that $\text{\hspace{0.17em}}g=h.\text{\hspace{0.17em}}$ We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse.

## Domain and range of inverse functions

The range of a function $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is the domain of the inverse function $\text{\hspace{0.17em}}{f}^{-1}\left(x\right).$

The domain of $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ is the range of $\text{\hspace{0.17em}}{f}^{-1}\left(x\right).$

Given a function, find the domain and range of its inverse.

1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.

## Finding the inverses of toolkit functions

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in [link] . We restrict the domain in such a fashion that the function assumes all y -values exactly once.

 Constant Identity Quadratic Cubic Reciprocal $f\left(x\right)=c$ $f\left(x\right)=x$ $f\left(x\right)={x}^{2}$ $f\left(x\right)={x}^{3}$ $f\left(x\right)=\frac{1}{x}$ Reciprocal squared Cube root Square root Absolute value $f\left(x\right)=\frac{1}{{x}^{2}}$ $f\left(x\right)=\sqrt[3]{x}$ $f\left(x\right)=\sqrt{x}$ $f\left(x\right)=|x|$

The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.

The absolute value function can be restricted to the domain $\text{\hspace{0.17em}}\left[0,\infty \right),$ where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain $\text{\hspace{0.17em}}\left(0,\infty \right).$

#### Questions & Answers

the sum of any two linear polynomial is what
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Momo
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6000
Robert
more than 6000
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can I see the picture
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with doing calculus
SLIMANE
Thanks po.
Jenica
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I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
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if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
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it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
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This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
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meena
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by how many trees did forest "A" have a greater number?
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32.243
Kenard