Lesson Finding Inverse

Source code of 'Finding Inverse'

This Lesson (Finding Inverse) was created by by Nate(3500)

Inverses of graphs are about finding symmetrical graphed lines along the identity function.
To find these inverse graphed lines, you need to switch the variables.
Example:
f(x) = 2x + 1
y = 2x + 1
x = 2y + 1
x - 1 = 2y
0.5x - 0.5 = y
0.5x - 0.5 = f^-1(x)
{{{graph( 600, 600, -10, 10, -10, 10, 2x + 1, 0.5x - 0.5, x ) }}}
If you see a patter for the points of the lines, you can see that the values are switched as well.
Points for f(x): (0,1), (1,3), and (2,5)
Points for f^-1(x): (1,0), (3,1), and (5,2)
Example:
f(x) = 1/x - 1
y = 1/x - 1
x = 1/y - 1
x + 1 = 1/y
y(x + 1) = 1
y = 1/(x + 1)
f^-1(x) = 1/(x + 1)
{{{graph( 600, 600, -10, 10, -10, 10, 1/x - 1, 1/(x + 1), x ) }}}
Example:
f(x) = x^2 - 1
y = x^2 - 1
x = y^2 - 1
x + 1 = y^2
+-sqrt(x + 1) = y
+-sqrt(x + 1) = f^-1(x)
{{{graph( 600, 600, -10, 10, -10, 10, x^2 - 1, sqrt(x + 1), -sqrt(x + 1), x ) }}}
Example:
f(x) = 1/(x^2 + 2)
y = 1/(x^2 + 2)
x = 1/(y^2 + 2)
(y^2 + 2)x = 1
y^2 + 2 = 1/x
y^2 = 1/x - 2
y = +-sqrt(1/x - 2)
f^-1(x) = +-sqrt(1/x - 2)
{{{graph( 600, 600, -10, 10, -10, 10, 1/(x^2 + 2), sqrt(1/x - 2), -sqrt(1/x - 2), x ) }}}
Example:
f(x) = sqrt(x^2) - 2
y = sqrt(x^2) - 2
x = sqrt(y^2) - 2
x + 2 = sqrt(y^2)
(x + 2)^2 = y^2
+-sqrt((x + 2)^2) = f^-1(x)
{{{graph( 600, 600, -10, 10, -10, 10, sqrt(x^2) + 2, sqrt((x - 2)^2), -sqrt((x - 2)^2), x ) }}}