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An image of two graphs. The first graph is of “y = f(x)”, which is a curved increasing function, that increases at a faster rate as x increases. The point (a, b) is on the graph of the function in the first quadrant. The second graph also graphs “y = f(x)” with the point (a, b), but also graphs the function “y = f inverse (x)”, an increasing curved function, that increases at a slower rate as x increases. This function includes the point (b, a). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
(a) The graph of this function f shows point ( a , b ) on the graph of f . (b) Since ( a , b ) is on the graph of f , the point ( b , a ) is on the graph of f −1 . The graph of f −1 is a reflection of the graph of f about the line y = x .

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 .

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from 0 to 2. The graph is of the function “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4).

Reflect the graph about the line y = x . The domain of f −1 is [ 0 , ) . The range of f −1 is [ −2 , ) . By using the preceding strategy for finding inverse functions, we can verify that the inverse function is f −1 ( x ) = x 2 2 , as shown in the graph.

An image of a graph. The x axis runs from -2 to 2 and the y axis runs from -2 to 2. The graph is of two functions. The first function is “f(x) = square root of (x +2)”, an increasing curved function. The function starts at the point (-2, 0). The x intercept is at (-2, 0) and the y intercept is at the approximate point (0, 1.4). The second function is “f inverse (x) = (x squared) -2”, an increasing curved function that starts at the point (0, -2). The x intercept is at the approximate point (1.4, 0) and the y intercept is at the point (0, -2). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
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Sketch the graph of f ( x ) = 2 x + 3 and the graph of its inverse using the symmetry property of inverse functions.


An image of a graph. The x axis runs from -3 to 4 and the y axis runs from -3 to 5. The graph is of two functions. The first function is “f(x) = 2x +3”, an increasing straight line function. The function has an x intercept at (-1.5, 0) and a y intercept at (0, 3). The second function is “f inverse (x) = (x - 3)/2”, an increasing straight line function, which increases at a slower rate than the first function. The function has an x intercept at (3, 0) and a y intercept at (0, -1.5). In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.

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Restricting domains

As we have seen, f ( x ) = 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 ( x ) = f ( x ) for all x in the domain of g . Then we can define an inverse function for g on that domain. For example, since f ( x ) = x 2 is one-to-one on the interval [ 0 , ) , we can define a new function g such that the domain of g is [ 0 , ) and g ( x ) = 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 ( x ) = x . On the other hand, the function f ( x ) = x 2 is also one-to-one on the domain ( , 0 ] . Therefore, we could also define a new function h such that the domain of h is ( , 0 ] and h ( x ) = 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 ( x ) = x ( [link] ).

An image of two graphs. Both graphs have an x axis that runs from -2 to 5 and a y axis that runs from -2 to 5. The first graph is of two functions. The first function is “g(x) = x squared”, an increasing curved function that starts at the point (0, 0). This function increases at a faster rate for larger values of x. The second function is “g inverse (x) = square root of x”, an increasing curved function that starts at the point (0, 0). This function increases at a slower rate for larger values of x. The first function is “h(x) = x squared”, a decreasing curved function that ends at the point (0, 0). This function decreases at a slower rate for larger values of x. The second function is “h inverse (x) = -(square root of x)”, an increasing curved function that starts at the point (0, 0). This function decreases at a slower rate for larger values of x. In addition to the two functions, there is a diagonal dotted line potted with the equation “y =x”, which shows that “f(x)” and “f inverse (x)” are mirror images about the line “y =x”.
(a) For g ( x ) = x 2 restricted to [ 0 , ) , g −1 ( x ) = x . (b) For h ( x ) = x 2 restricted to ( , 0 ] , h −1 ( x ) = x .

Restricting the domain

Consider the function f ( x ) = ( x + 1 ) 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 [ −1 , ) . 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.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, which is a parabola. The function decreases until the point (-1, 0), where it begins it increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1). There is also a horizontal dotted line plotted on the graph, which crosses through the function at two points.
  2. On the interval [ −1 , ) , f is one-to-one.
    An image of a graph. The x axis runs from -6 to 6 and the y axis runs from -2 to 10. The graph is of the function “f(x) = (x+ 1) squared”, on the interval [1, infinity). The function starts from the point (-1, 0) and increases. The x intercept is at the point (-1, 0) and the y intercept is at the point (0, 1).
    The domain and range of f −1 are given by the range and domain of f , respectively. Therefore, the domain of f −1 is [ 0 , ) and the range of f −1 is [ −1 , ) . To find a formula for f −1 , solve the equation y = ( x + 1 ) 2 for x . If y = ( x + 1 ) 2 , then x = −1 ± y . Since we are restricting the domain to the interval where x −1 , we need ± y 0 . Therefore, x = −1 + y . Interchanging x and y , we write y = −1 + x and conclude that f −1 ( x ) = −1 + x .
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Consider f ( x ) = 1 / x 2 restricted to the domain ( , 0 ) . 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 ( 0 , ) . The range of f −1 is ( , 0 ) . The inverse function is given by the formula f −1 ( x ) = −1 / x .

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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 [ π 2 , π 2 ] . By doing so, we define the inverse sine function on the domain [ −1 , 1 ] such that for any x in the interval [ −1 , 1 ] , the inverse sine function tells us which angle θ in the interval [ π 2 , π 2 ] satisfies sin θ = 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.

Questions & Answers

can you give me a problem for function. a trigonometric one
geovanni Reply
state and prove L hospital rule
Krishna Reply
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
Jawad
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were $48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach $60 billion?
Nguyen Reply
what is the derivative of x= Arc sin (x)^1/2
morfling Reply
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Obi Reply
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
Ryan Reply
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)
NIKI Reply
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
K.kupar Reply
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
Darnell Reply
proof the formula integration of udv=uv-integration of vdu.?
Bg Reply
Find derivative (2x^3+6xy-4y^2)^2
Rasheed Reply
no x=2 is not a function, as there is nothing that's changing.
Vivek Reply
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
Joys Reply
y=800
Gift
800
Bg
how do u factor the numerator?
Drew Reply
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
follow algebraic method. look under factoring numerator from Khan academy
moe
volume between cone z=√(x^2+y^2) and plane z=2
Kranthi Reply
answer please?
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
Practice Key Terms 5

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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