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Question: Let \(f(x) = {x^2} - 16\) how do you find \({f^{ - 1}}(x)?\)...

Let f(x)=x2−16f(x) = {x^2} - 16 how do you find f−1(x)?{f^{ - 1}}(x)?

Explanation

Solution

To find f−1(x){f^{ - 1}}(x) or inverse of function f(x)f(x), first replacef(x)f(x) with yy from the given equation and then switch the variables x  and  yx\;{\text{and}}\;y with each other then solve the switched equation for yy and then finally replace yy with f−1(x){f^{ - 1}}(x), you will get the required inverse function of the given function.

Complete step by step answer:
In order to find the inverse of the given function f(x)=x2−16f(x) = {x^2} - 16 we will first replace f(x)f(x) with yy, that is
f(x)=x2−16 ⇒y=x2−16 f(x) = {x^2} - 16 \\\ \Rightarrow y = {x^2} - 16 \\\
Again we will switch x  and  yx\;{\text{and}}\;y with each other, we will get

\Rightarrow y = {x^2} - 16 \\\ \Rightarrow x = {y^2} - 16 \\\ $$ Now we will solve this equation for the variable $y$ To solve above equation for $y$ adding $16$ to the left hand side and the right hand side both, we will get

\Rightarrow x = {y^2} - 16 \\
\Rightarrow x + 16 = {y^2} - 16 + 16 \\
\Rightarrow x + 16 = {y^2} \\ $$
Now taking the square root both sides we will get,

\Rightarrow x + 16 = {y^2} \\\ \Rightarrow \sqrt {x + 16} = \sqrt {{y^2}} \\\ \Rightarrow \sqrt {x + 16} = y \\\ \Rightarrow y = \sqrt {x + 16} \\\ $$ Finally replacing $y\;{\text{with}}\;{f^{ - 1}}(x)$ $$ \therefore {f^{ - 1}}(x) = \sqrt {x + 16} $$ **Therefore the required inverse function of the function $f(x) = {x^2} - 16$ is given as $${f^{ - 1}}(x) = \sqrt {x + 16} $$.** **Note:** Actually inverse function for the function $f(x) = {x^2} - 16$ does not exist because only one to one or one-one functions have their inverse function whereas $f(x) = {x^2} - 16$ is not a one to one function because it has more than one argument for a single value or output of the function. So you may be thinking of if its inverse function does not exist then what have does in the solution, actually that is the inverse function of the function $f(x) = {x^2} - 16$ with its domain restricted to the interval $\left[ {0,\;\infty } \right]$ and by restricting the domain we have made the function one to one.