Examples from Section 6.1
Inverse Functions and Their Derivatives

p. 654

Identifying one-to-one functions

3.

The function tan (x) is not one-to-one because it fails the horizontal line test. However, if the domain is restricted to the interval [-π/2,π/2], then tan (x) is one-to-one and has the (standard) inverse arctan (x).

Graphing Inverse Functions

7.

The eaisest way to grpah inverse functions is to use ParametricPlot[], switching the x and y coordinates.

p  =  
      ParametricPlot[{x, 1/(1 + x^2)},  {x, 0, 2},  
        PlotRange -> {0, 2}, PlotStyle -> {Hue[ .3]}]

q  =  
      ParametricPlot[{1/(1 + x^2),  x},  {x, 0, 2},  
        PlotRange -> {{0, 1}, {0, 2}},  PlotStyle -> {Hue[ .7]}]

Show[p, q]

Formulas for Inverse Functions

17.

Find a formula for the inverse of f (x)  =  (((x + 1))^2) for x  ≥  -1.

f[x_]  =  (x + 1)^2;

Solve[ y  ==  f[x], x]

There are two "branches" for the inverse. Since we are restricting to x  ≥  -1 we choose x  ->  -1  +  (Sqrt[y]) with y  ≥  0. As a function of x the inverse would be written:

invf[x_]  =  -1  +  (Sqrt[x]);

p  =  Plot[f[x], {x, -1, 2},  PlotStyle -> Hue[ .3]]

q  =  Plot[invf[x],  {x, 0, 9}, PlotStyle -> Hue[ .7]]

Show[p, q]

Derivatives of Inverse Functions

29.

a)

f[x_]  =  (x^3);  invf[x]  =  (x^(1/3));

f[invf[x]]

b) In order to force Mathematica to work with real numbers only (i.e., to ingore complex valued solutions) we invoke the following package:

Needs["Miscellaneous`RealOnly`"]

Plot[{f[x], invf[x]}, {x, -1.5, 1.5},  
      PlotStyle -> {Hue[ .3],  Hue[ .7]}]

c) Slopes of tangent lines to f (x)  =  (x^3)at -1 and 1:

f[x_]  :=  (x^3);  f'[x]

f'[-1]

f'[1]

c) Slopes of tangent lines to invf (x)  =  x^(1/3)at -1 and 1:

invf[x_]  :=  (x^(1/3));  invf'[x]

invf'[-1]

invf'[1]

d) The slope of f (x)  =  (x^3) at 0 is 0 so the tangent line is the horizontal x-axis. Thus, the tangent line for the inverse function invf[x]  =  (x^(1/3)) is the vertical y-axis; its slope is "infinite":

invf'[0]

CAS Explorations and Projects

56.

f (x) =  (x^3)/((x^2) + 1), -1  ≤  x  ≤  1, (Subscript[x, 0])  =  1/2.
a) f (x) is one-to-one

Clear[x, y]

f[x_]  :=  (x^3)/((x^2) + 1);

Plot[{f[x], f'[x]},  {x, -1, 1},  
      PlotStyle -> {Hue[ .3], Hue[ .7]}, 
      PlotRange -> {{-1, 1}, {-1, 1}}]

The derivative (blue) is always positive (and 0 only at 0) so the function is strictly increasing on the interval [-1,1] and consequently has an inverse there.

b) Formula for inverse.

Solve[ x  ==  f[y],  y]

Substitute the third solution for g (x):

g[x_]  :=  y  /.  %[[3]]

c) Equation of the tangent line of f (x) at (1/2, f (1/2)).

f[1/2]

f'[1/2]

y  ==  1/10  +  13/25 (x  -  1/2)

d) Equation of the tangent line of g (x) at (f (1/2), 1/2). By Theorem 1, the slope is 1/f' (1/2)  =  25/13. So the tangent line is

y  ==  1/2  +  25/13 (x  -  1/10)

e) The line connecting (1/2, 1/10) and (1/10, 1/2) is given by

y  ==  
      1/10  +  
        (1/2  -  1/10)/(1/10  -  1/2) (x  -  1/2)

\(Plot[{f[x], 1/10\  + \ 13/25\ \((x\  - \ 1/2)\), \ \ g[x], \ 
        1/2\  + \ 25/13\ \((x\  - \ 1/10)\), \ x, \ 3\/5 - x}, 
      \ {x, \(-1\), 1}, \ PlotRange -> {\(-1.5\), 1.5}, 
      PlotStyle -> {RGBColor[0, 0, 1], RGBColor[0, 1, 1], \ 
          RGBColor[0, 1, 0], \ RGBColor[1, 1, 0], \ GrayLevel[ .75], \ 
          GrayLevel[ .75]}, 
AspectRatio -> Automatic]\)