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 $\text{\hspace{0.17em}}\left(0,\infty \right).$

The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
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
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
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).
Natalie
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
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
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
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations