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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

f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
pweding paturo nsa calculus?
jimmy
how to use fundamental theorem to solve exponential
JULIA Reply
find the bounded area of the parabola y^2=4x and y=16x
Omar Reply
what is absolute value means?
Geo Reply
Chicken nuggets
Hugh
🐔
MM
🐔🦃 nuggets
MM
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
Game Reply
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
Godwin Reply
how does this work
Brad Reply
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
Cosmos Reply
log tan (x/4+x/2)
Rohan
please answer
Rohan
y=(x^2 + 3x).(eipix)
Claudia
is this a answer
Ismael
A Function F(X)=Sinx+cosx is odd or even?
WIZARD Reply
neither
David
Neither
Lovuyiso
f(x)=1/1+x^2 |=[-3,1]
Yuliana Reply
apa itu?
fauzi
determine the area of the region enclosed by x²+y=1,2x-y+4=0
Gerald Reply
Hi
MP
Hi too
Vic
hello please anyone with calculus PDF should share
Adegoke
Which kind of pdf do you want bro?
Aftab
hi
Abdul
can I get calculus in pdf
Abdul
explain for me
Usman
okay I have such documents
Fitzgerald
please share it
Hamza
How to use it to slove fraction
Tricia Reply
Hello please can someone tell me the meaning of this group all about, yes I know is calculus group but yet nothing is showing up
Shodipo
You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
Jean
Im not in college but this will still help
nothing
how en where can u apply it
Migos
how can we scatch a parabola graph
Dever Reply
Ok
Endalkachew
how can I solve differentiation?
Sir Reply
with the help of different formulas and Rules. we use formulas according to given condition or according to questions
CALCULUS
For example any questions...
CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
Rohan
what is the procedures in solving number 1?
Vier Reply
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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