Chapter 6: Inverse Functions


A function f(x)f(x) is ONE-TO-ONE on a domain DD if f(x1)f(x2)f(x_1) \neq f(x_2)  Whenever x1x2x_1 \neq x_2  in DD

Plainly f(x)f(x) is one-to-one if every xx gets sent to a different value


The Horizontal Line Test - A function is one-to-one if and only if each horizontal line meets it graph at most once.


Which of these functions is one-to-one ?

  1. y=xy = |x| 
  2. y=x3y= x^3 
  3. y=3x+2y=3x+2
  4. y=sin(x)y= \sin (x) 
  5. y=exy =e^x 
  6. y=x2y= x^2


If f(x)f(x) is one-to-one on DD and has range RR then the INVERSE FUNCTION is defined by:

f1(b)=af^{-1}(b) = a if f(a)=bf(a) =b

Why do we need one-to-one?

If f(1)=f(2)=3f (1) = f(2) =3 then what is f1(3)f^{-1}(3) ?

It could be f1(3)=1f^{-1}(3) = 1 or f1(3)=2f^{-1}(3) =2

For f1(x)f^{-1}(x)  to be a function we need one output

The Inversion Procedure

To invert f(x)f(x)

Solve y=f(x)y= f(x) for xx. Obtain x=f1(y)x = f^{ -1}(y)

Switch xx and yy to obtain y=f1(x)y = f^{-1}(x).


Invert the function y=12x+1y = \frac{1}{2} x+ 1

y=12x+12y=x+2y = \frac{1}{2} x +1 \Leftrightarrow 2y = x+2

2y2=x\Leftrightarrow 2y -2 =x

Thus, x=f1(y)=2y2x= f^{-1}(y) = 2y-2

We obtain f1(x)=2x2f^{-1}(x) = 2x-2

* Check your work

f(f1(x)=f(2x2)=12(2x2)+1f(f^{-1}(x) = f(2x-2) = \frac{1}{2}(2x-2)+1

== x1+1=x x-1+1 =x

f1(f(x))=2(12x+1)2=x+22f^{-1}(f(x))=2 (\frac{1}{2} x +1) -2 = x +2 -2



Invert the function y=x2y=x^2 on the domain [0,)[0, \infty).

* y=x2y=x^2  is not one-to-one on (,)(- \infty , \infty) but it is on [0,)[0, \infty).

y=x2y=xx=f1(y)=yy = x ^2 \Rightarrow \sqrt{y} = x \Rightarrow x = f ^{-1} (y) = \sqrt{y}

We obtain f1(x)=x f ^{-1} (x) = \sqrt{x}


f(f1(x))=(x)2=xf(f^{-1}(x)) = (\sqrt{x})^2 = x

f1(f(x))=x2=x=xf^{-1}(f(x)) = \sqrt{x^2} = |x|=x

Inverse Trigonometric Functions

Idea: The trig functions are NOT one-to-one.

0=sin(0)=sin(2π)=sin(4π)=...0 = \sin (0) = \sin (2 \pi) = \sin (4 \pi) = ...

However, we want to invest them

Domain Restrictions

We get the following inverses:

y=arcsin(x)xy = \arcsin (x) \rightarrow x in [1,1][-1,1]

y=arccos(x)xy = \arccos (x) \rightarrow x in [1,1][-1,1] 

y=arctan(x)xy = \arctan (x) \rightarrow x in (,)( - \infty, \infty)


What are the ranges of:

arcsinx,arccosx\arcsin x, \arccos x and arctanx\arctan x

Fun Fact


Calculate arcsin(32)\arcsin (\frac{\sqrt{3}}{2})

Find θ\theta in [π2, π2][- \frac {\pi}{2}, \ \frac {\pi}{2}] such that sinθ=32\sin \theta = \frac {\sqrt{3}}{2}

We know sinπ6=32\sin \frac {\pi}{6} = \frac {\sqrt{3}}{2} since

Thus arcsin(32)=π6\arcsin (\frac{\sqrt{3}}{2})= \frac{\pi}{6}


Calculate arccos(12)\arccos (-\frac{1}{2})

Find θ\theta in [0,π][0,\pi]  such that cosθ=12\cos \theta = - \frac{1}{2}

We know cos(2π3)=12\cos (\frac{2 \pi}{3}) = -\frac{1}{2}  since

Thus , arccos(12)=2π3\arccos(- \frac {1}{2}) = \frac {2 \pi}{3}


Sketch y=arcsin(x)y = \arcsin (x)

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