<< Chapter < Page Chapter >> Page >
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”.
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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”.

Got questions? Get instant answers now!

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 .
Got questions? Get instant answers now!
Got questions? Get instant answers now!

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 .

Got questions? Get instant answers now!

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

The f'(4)for f(x) =4^x
Alice Reply
I need help under implicit differentiation
Uchenna Reply
how to understand this
nhie
understand what?
Boniface
how dont you understand it is obvious
Se
if it's there, it must be differentiable
Giorgio
marginal rate of substitution in economics for an example is an implicit differentiation function. It is a proportion of comparison. It could be expressed as the area of a triangle of the old value proportional to the new, and then the next value and so on.
James
The new shapes form a line with a derivative curve
James
That curve could be expressed mathematically. A good real life example is the proportion at which people barter in pawn shops. "How about 100?" "What are you trying to do? rob me? 50!" "No way chap. 75." "Ill give you 62. deal?" "68. Nothing more nothing less." "deal"
James
the proportion of what someone was trying to get for their product, versus what they were offered to the new price they wanted for their product and what they were offered
James
The proportion of the differentiating triangles would be somewhat 1:2. and since there is little variation to the curve then it looks more like a straight line.
James
By the way This is one of the hardest subjects for me. I have a really hard time expressing things in such a way. I'm trying to have more exact calculations which is why I still study the subject.
James
What is derivative of antilog x dx ?
Tanmay Reply
what's the meaning of removable discontinuity
Brian Reply
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
please help me to calculus
World Reply
may god be with you
Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question? or ANY of the question?
Levis Reply
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
MOHAMMAD
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
Levis Reply
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
Michagaye Reply
i have not understood
Leo
The answer is 0
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
ha hakdog hahhahahaha
No Reply
ha hamburger
Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
albert Reply
What's f(x) ^x^x
Emeka Reply
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
if your asking for derivative dy/dz=x^2/2(lnx-1/2)
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
So in that case what will be the answer?
Alice
nice explanation Levis, appreciated..
Thato
what is a maximax
Chinye Reply
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
Dike Reply
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
Care Reply
3 + 3=6
mujahid
How to do basic integrals
dondi Reply
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
Practice Key Terms 5

Get the best Calculus volume 1 course in your pocket!





Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask