# 1.4 Inverse functions  (Page 3/11)

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## Sketching graphs of inverse functions

For the graph of $f$ in the following image, sketch a graph of ${f}^{-1}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of ${f}^{-1}.$

Reflect the graph about the line $y=x.$ The domain of ${f}^{-1}$ is $\left[0,\infty \right).$ The range of ${f}^{-1}$ is $\left[-2,\infty \right).$ By using the preceding strategy for finding inverse functions, we can verify that the inverse function is ${f}^{-1}\left(x\right)={x}^{2}-2,$ as shown in the graph.

Sketch the graph of $f\left(x\right)=2x+3$ and the graph of its inverse using the symmetry property of inverse functions.

## Restricting domains

As we have seen, $f\left(x\right)={x}^{2}$ does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of $f$ such that the function is one-to-one. This subset is called a restricted domain    . By restricting the domain of $f,$ we can define a new function $g$ such that the domain of $g$ is the restricted domain of $f$ and $g\left(x\right)=f\left(x\right)$ for all $x$ in the domain of $g.$ Then we can define an inverse function for $g$ on that domain. For example, since $f\left(x\right)={x}^{2}$ is one-to-one on the interval $\left[0,\infty \right),$ we can define a new function $g$ such that the domain of $g$ is $\left[0,\infty \right)$ and $g\left(x\right)={x}^{2}$ for all $x$ in its domain. Since $g$ is a one-to-one function, it has an inverse function, given by the formula ${g}^{-1}\left(x\right)=\sqrt{x}.$ On the other hand, the function $f\left(x\right)={x}^{2}$ is also one-to-one on the domain $\left(\text{−}\infty ,0\right].$ Therefore, we could also define a new function $h$ such that the domain of $h$ is $\left(\text{−}\infty ,0\right]$ and $h\left(x\right)={x}^{2}$ for all $x$ in the domain of $h.$ Then $h$ is a one-to-one function and must also have an inverse. Its inverse is given by the formula ${h}^{-1}\left(x\right)=\text{−}\sqrt{x}$ ( [link] ).

## Restricting the domain

Consider the function $f\left(x\right)={\left(x+1\right)}^{2}.$

1. Sketch the graph of $f$ and use the horizontal line test to show that $f$ is not one-to-one.
2. Show that $f$ is one-to-one on the restricted domain $\left[-1,\infty \right).$ Determine the domain and range for the inverse of $f$ on this restricted domain and find a formula for ${f}^{-1}.$
1. The graph of $f$ is the graph of $y={x}^{2}$ shifted left 1 unit. Since there exists a horizontal line intersecting the graph more than once, $f$ is not one-to-one.
2. On the interval $\left[-1,\infty \right),\phantom{\rule{0.2em}{0ex}}f$ is one-to-one.

The domain and range of ${f}^{-1}$ are given by the range and domain of $f,$ respectively. Therefore, the domain of ${f}^{-1}$ is $\left[0,\infty \right)$ and the range of ${f}^{-1}$ is $\left[-1,\infty \right).$ To find a formula for ${f}^{-1},$ solve the equation $y={\left(x+1\right)}^{2}$ for $x.$ If $y={\left(x+1\right)}^{2},$ then $x=-1±\sqrt{y}.$ Since we are restricting the domain to the interval where $x\ge -1,$ we need $±\sqrt{y}\ge 0.$ Therefore, $x=-1+\sqrt{y}.$ Interchanging $x$ and $y,$ we write $y=-1+\sqrt{x}$ and conclude that ${f}^{-1}\left(x\right)=-1+\sqrt{x}.$

Consider $f\left(x\right)=1\text{/}{x}^{2}$ restricted to the domain $\left(\text{−}\infty ,0\right).$ Verify that $f$ is one-to-one on this domain. Determine the domain and range of the inverse of $f$ and find a formula for ${f}^{-1}.$

The domain of ${f}^{-1}$ is $\left(0,\infty \right).$ The range of ${f}^{-1}$ is $\left(\text{−}\infty ,0\right).$ The inverse function is given by the formula ${f}^{-1}\left(x\right)=-1\text{/}\sqrt{x}.$

## Inverse trigonometric functions

The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ( [link] ). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right].$ By doing so, we define the inverse sine function on the domain $\left[-1,1\right]$ such that for any $x$ in the interval $\left[-1,1\right],$ the inverse sine function tells us which angle $\theta$ in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$ satisfies $\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =x.$ Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions    , which are functions that tell us which angle in a certain interval has a specified trigonometric value.

what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The