Inverse Functions

Related to Section 6.1 in Stewart, 8th edition

Overview

We need and use a number of different kinds of functions:

Linear functions f(x)=m x+b are the simplest functions. We can approximate many of the simplest behavior with these “straight line” functions.

Plot[3x+5,{x,-2,1}]

Quadratic functions f(x)=a (x^2)+b x+c are useful to model things like the effects of gravity and shapes used in satellite dishes and headlights (“parabola” or “paraboloid”).

Plot[-9x^2+3x+5,{x,-2,1}]

Polynomial functions are used to approximate complicated phenomena. For instance piecewise cubic polynomials (“splines”) are used in drawing applications like Adobe Illustrator.

Plot[If[x<0,3x+5,-9x^2+3x+5],{x,-2,1}]

Sinusoidal functions with sines and cosines model repeating (“periodic”) phenomena.

Plot[3Sin[Pi x]+5,{x,-2,1}]

Questions

Solve the following. (In these expressions x and y are called variables, whereas a and b are called parameters. Think of a parameter as an unspecified yet fixed number.)

Solve the linear equation y=3x+5=b for x. (Solution)

So you might think to yourself,
“When y takes the value b, what is the value of x?”

Solve the quadratic equation y=((x^2)/2)-b=1 for x. (Solution)

Solve[x^2/2-b==1,x]

Solve the cubic equation y=2(x^3)-3=a for x. (Solution)

Solve the sinusoidal equation y=2sin(x)=1 for x. (Solution)

Solve[2Sin[x]==1,x]
Plot[{2Sin[x],1},{x,0,2Pi}]

What’s troubling about the

quadratic inverse problem?

sinusoidal inverse problem?

Inverse functions

Inverse functions is a framework for solving equations. We have “obvious” methods for solving some of them, e.g.

Use square roots to solve quadratic equations.

I’m fond of that student who, when asked for the square root of 4, says “-2”. It’s true! But we’re trained to ignore that “solution”, and take the positive root:

Use cube roots to solve simple cubic equations.

Same deal for cubics:

Definition

Given a function f(x): if y=f(x) has exactly one solution x for any  y in the range of f, then the inverse to f exists, and is denoted (f^(-1)):
			y=f(x) ⟷ (f^(-1))(y)=x
We say that f is “one-to-one”; we pronounce “⟷” as “if and only if”.

Example

If f(x)=2(x^3)-3, then to find (f^(-1)) solve
	y=f(x)=2(x^3)-3 for x:
	⟶ y+3=2(x^3)
	⟶ ((y+3)/2)=(x^3)
	⟶ x=(((y+3)/2)^(1/3))

The inverse function to f(x)=2(x^3)-3 is (f^(-1))(y)=(((y+3)/2)^(1/3)) or
	(f^(-1))(x)=(((x+3)/2)^(1/3))

Question

If g(x)=((x^2)/2)+1 find (g^(-1))(x), if it exists. (Solution)

Plot[{x^2/2+1,3},{x,-3,3},PlotStyle->{Blue,Dashed}]

Inverse function properties

A function f is one-to-one if it never takes on the same value twice, i.e. f((Subscript[x, 1]))≠f((Subscript[x, 2])) whenever (Subscript[x, 1])≠(Subscript[x, 2]). One-to-one functions have exactly one inverse.

A horizontal line can cross the graph of a one-to-one function at most once.

Question

Which is the graph of a one-to-one function? If it is not a one-to-one function, how can the domain be restricted to make a one-to-one function? (Solution)

GraphicsGrid[{{Plot[(2^x),{x,-4,4},PlotLabel->"Plot 1"],Plot[1/x,{x,-10,10},PlotLabel->"Plot 2"]},{Plot[5 (E^(-(((x-1))^2)/10)),{x,-10,10},PlotLabel->"Plot 3"],Plot[((x^3)-16x)/5,{x,-10,10},PlotRange->{-10,10},PlotLabel->"Plot 4"]}}]

GraphicsGrid[{{Plot[{(2^x),3},{x,-4,4},PlotLabel->"Plot 1"],Plot[{1/x,3},{x,-10,10},PlotLabel->"Plot 2"]},{Plot[{5 (E^(-(((x-1))^2)/10)),3},{x,-10,10},PlotLabel->"Plot 3"],Plot[{((x^3)-16x)/5,3},{x,-10,10},PlotRange->{-10,10},PlotLabel->"Plot 4"]}}]

Properties

(f^(-1))(f(x))=x for every x in the domain of f.

f((f^(-1))(y))=y for every y in the domain of (f^(-1)).

(Notice the change in variable above: 
if f takes values of x to y, then (f^-1) takes values of y to x. 
Nonetheless, we typically use x to represent the dependent variable in a function.

The domain of (f^(-1)) is the same as the range of f, and the range of (f^(-1)) is the same as the domain of f.

Since y=f(x) is the same as (f^(-1))(y)=x, the graph y=(f^(-1))(x) is the same as the graph y=f(x) except x and y have been exchanged: i.e. the graph has been reflected about the line y=x.

Questions

Each question refers to the function in Plot 1:

Plot[(2^x),{x,-4,4},PlotLabel->"Plot 1",AspectRatio->1]

What is the graph of the inverse function? (Solution)

Plot[{(2^x),Log[2,x],x},{x,-4,16},PlotLabel->"Plot 1",AspectRatio->1,PlotRange->{-4,18},PlotStyle->{Blue,Black,Dashed}]

What is the domain of the inverse function? (Solution)

Estimate the value of the inverse function (f^(-1))(x) when x=4. (Solution)

Calculus of inverses

Use the fact that y=(f^(-1))(x) is the same as f(y)=x, and implicit differentiation to find the derivative of (f^(-1))(x).

Example

We know what the derivative of (x^(1/3)), but we can also get it from the cube function:
	y=(x^(1/3)) ⟶x=(y^3) 
	⟶ (d/(d x))(x)=(d/(d x))((y^3))
	⟶ 1=3(y^2)((d y)/(d x))
	⟶ ((d y)/(d x))=(1/(3(y^2)))=(1/(3((((x^(1/3))))^2)))

Question

Apply this reasoning to find a formula for (d/(d x))(f^(-1))(x). (Solution)

Homework

IMath problems on inverse functions