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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

what is hyperbolic function
vector Reply
find volume of solid about y axis and y=x^3, x=0,y=1
amisha Reply
3 pi/5
vector
what is the power rule
Vanessa Reply
how do i deal with infinity in limits?
Itumeleng Reply
Add the functions f(x)=7x-x g(x)=5-x
Julius Reply
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
Cabdalla Reply
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
mahin Reply
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
riyad Reply
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
abigail Reply
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
Rodgers Reply
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
Sudhanayaki Reply
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
Crystal Reply
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Robert Reply
can you please post the question again here so I can see what your talking about
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
Robert Reply
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
aliyu Reply
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
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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