An inverse function is a function such that [math]f^{-1}\left(f\left(x\right)\right)=x[/math] and [math]f\left(f^{-1}\left(x\right)\right)=x[/math]. In other words, given a function f(x), which has some operations(squaring, multiplying, adding..), the inverse function [math]f^{-1}(x)[/math] will perform the opposite operations of the original function(square root, dividing, subtracting...) in such a way that when evaluating [math]f^{-1}\circ f[/math] or [math]f\circ f^{-1}[/math], the original value, x, will be obtained.[br][br]The graph below shows the function [math]f\left(x\right)=e^x[/math] and two points: [math]A=(a,e^a)[/math] and [math]B=(e^a,a)[/math]. By adjusting the values of [i]a[/i], you can move the points A and B. Notice how they move along the graph: as A moves along [math]f(x)=e^x[/math], B moves along [math]g(x)=ln(x)[/math].[br][br]For example, when [i]a=1[/i], [math]f(1)=e^1=e[math]\approx[/math]2.72[/math] and [math]g(2.72)=ln(2.72)=ln(e)=1[/math]. When we put the value 1 in[math]f(x)[/math], it returns [math]e[/math], after that we put [math]e[/math] into [math]g(x)[/math] and it returns to us 1, this process is the same as performing [math]g\left(f\left(1\right)\right)=g\circ f=ln\left(e^1\right)=1[/math], and since the returned value is the original value, [math]g(x)=f^{-1}[/math]is the inverse function of [math]f(x)[/math]

Using the graph of [math]f(x)=e^x[/math], and your knowledge of inverse functions, can you estimate the values of [math]ln(1),ln(0.5),ln(2)[/math]? What about [math]ln(e)[/math]?

To have an inverse, a function must satisfy two requirements: be injective(one-to-one) and surjective(onto), i.e. it must be a [i]bijection[/i]. The graph of the inverted function will always be the mirror image of the original function reflected about the y=x line.[br][br]In the graph below you can see this reflection in action. By adjusting the values of [i]a[/i], the point [math]G[/math]moves along the graph [math]g(x)=e^x[/math] and [math]G_1[/math] traces the graph of the inverse. The same happens for [math]h\left(x\right)=\frac{\left(x^3+1\right)}{3}[/math] and the points [math]H, H_1[/math]. You can click on the circles next to the functions/points to hide/show them.

The point [math]F[/math] draws the graph of a function [math]f\left(x\right)[/math] as you change [i]a[/i], find [math]f\left(x\right)[/math] and [math]f^{-1}\left(x\right)[/math] using the values of [math]F[/math]. Then, click on the empty box under the points and type in your function and inverse, see if they are reflected about the y=x line.

It becomes clear why functions that are not bijections cannot have an inverse simply by analysing their graphs. The example below shows the graph of [math]f(x)=x^2[/math] and its reflection along the y=x line. Why is the reflection not the inverse function of [math]f(x)[/math]? (tip: recall the vertical line test)