When we think about the graphs of inverse functions, geometrically we are talking about reflecting the function $f(x)$f(x) over the line $y=x$y=x to draw the inverse function $f^{-1}\left(x\right)$f−1(x).
By doing this the $x$x values, or the inputs, become the $y$y values, or the outputs, and vice versa.
Let's start by reflecting a few points, belonging to a curve, over the line $y=x$y=x to see this in action.
We can see that $(0,2)$(0,2) is reflected across to $(2,0)$(2,0).
Similarly $(1,3)$(1,3) is transformed to $(3,1)$(3,1) and $(2,6)$(2,6) to $(6,2)$(6,2).
So when we want to use the graph of $f(x)$f(x) to draw its inverse $f^{-1}\left(x\right)$f−1(x), then we want to take points on the curve of $f(x)$f(x) and reflect them over the line $y=x$y=x.
Let's take a look at an example.
example
The function $f\left(x\right)=\sqrt{x+3}+1$f(x)=√x+3+1 is graphed below along with the line $y=x$y=x. Sketch the graph of $f^{-1}\left(x\right)$f−1(x).
Think: We'll first identify some points on $f(x)$f(x) and then reflect them over $y=x$y=x.
Do:
Worked Examples
example 1
Below we have sketched the line $y=\frac{1}{2}x$y=12x (labelled $B$B) over the line $y=x$y=x (labelled $A$A).
By reflecting $y=\frac{1}{2}x$y=12x about the line $y=x$y=x, graph the inverse of $y=\frac{1}{2}x$y=12x.
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example 2
Below we have sketched the line $y=\frac{x^2}{4}+1$y=x24+1 as defined for $x\le0$x≤0 (labelled $B$B) over the line $y=x$y=x (labelled $A$A).
By reflecting this arm of $y=\frac{x^2}{4}+1$y=x24+1 about the line $y=x$y=x, graph the inverse of the arm of $y=\frac{x^2}{4}+1$y=x24+1 defined over $x\le0$x≤0.
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example 3
Below we have graphed the line $y=\left(\frac{3}{2}\right)^{-x}$y=(32)−x (labelled $B$B) over the line $y=x$y=x (labelled $A$A).
By reflecting this arm of $y=\left(\frac{3}{2}\right)^{-x}$y=(32)−x about the line $y=x$y=x, graph the inverse of the arm of $y=\left(\frac{3}{2}\right)^{-x}$y=(32)−x.
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