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For the graph of $f$ in the following image, sketch a graph of ${f}^{\mathrm{-1}}$ by sketching the line $y=x$ and using symmetry. Identify the domain and range of ${f}^{\mathrm{-1}}.$
Reflect the graph about the line $y=x.$ The domain of ${f}^{\mathrm{-1}}$ is $\left[0,\infty \right).$ The range of ${f}^{\mathrm{-1}}$ is $\left[\mathrm{-2},\infty \right).$ By using the preceding strategy for finding inverse functions, we can verify that the inverse function is ${f}^{\mathrm{-1}}\left(x\right)={x}^{2}-2,$ as shown in the graph.
Sketch the graph of $f\left(x\right)=2x+3$ and the graph of its inverse using the symmetry property of inverse functions.
As we have seen, $f\left(x\right)={x}^{2}$ does not have an inverse function because it is not one-to-one. However, we can choose a subset of the domain of $f$ such that the function is one-to-one. This subset is called a restricted domain . By restricting the domain of $f,$ we can define a new function $g$ such that the domain of $g$ is the restricted domain of $f$ and $g(x)=f(x)$ for all $x$ in the domain of $g.$ Then we can define an inverse function for $g$ on that domain. For example, since $f(x)={x}^{2}$ is one-to-one on the interval $[0,\infty ),$ we can define a new function $g$ such that the domain of $g$ is $[0,\infty )$ and $g(x)={x}^{2}$ for all $x$ in its domain. Since $g$ is a one-to-one function, it has an inverse function, given by the formula ${g}^{\mathrm{-1}}(x)=\sqrt{x}.$ On the other hand, the function $f(x)={x}^{2}$ is also one-to-one on the domain $(\text{\u2212}\infty ,0].$ Therefore, we could also define a new function $h$ such that the domain of $h$ is $(\text{\u2212}\infty ,0]$ and $h(x)={x}^{2}$ for all $x$ in the domain of $h.$ Then $h$ is a one-to-one function and must also have an inverse. Its inverse is given by the formula ${h}^{\mathrm{-1}}(x)=\text{\u2212}\sqrt{x}$ ( [link] ).
Consider the function $f\left(x\right)={\left(x+1\right)}^{2}.$
Consider $f\left(x\right)=1\text{/}{x}^{2}$ restricted to the domain $\left(\text{\u2212}\infty ,0\right).$ Verify that $f$ is one-to-one on this domain. Determine the domain and range of the inverse of $f$ and find a formula for ${f}^{\mathrm{-1}}.$
The domain of ${f}^{\mathrm{-1}}$ is $\left(0,\infty \right).$ The range of ${f}^{\mathrm{-1}}$ is $(\text{\u2212}\infty ,0).$ The inverse function is given by the formula ${f}^{\mathrm{-1}}\left(x\right)=\mathrm{-1}\text{/}\sqrt{x}.$
The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function ( [link] ). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict the domain to the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$ By doing so, we define the inverse sine function on the domain $[\mathrm{-1},1]$ such that for any $x$ in the interval $[\mathrm{-1},1],$ the inverse sine function tells us which angle $\theta $ in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ satisfies $\text{sin}\phantom{\rule{0.1em}{0ex}}\theta =x.$ Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric functions , which are functions that tell us which angle in a certain interval has a specified trigonometric value.
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