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

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Answer 1 · 2018-03-21T15:04:45

This is a way to express finding the inverse function of $f (x) = x^{2} - 16$ $f(x)=x^2-16$

First, write the function as $y = x^{2} - 16$ $y=x^2-16$ .

Next, switch the $y$ $y$ and $x$ $x$ positions.

$x = y^{2} - 16 \to$ $x=y^2-16 rarr$ Solve for $y$ $y$ in terms of $x$ $x$

$x + 16 = y^{2}$ $x+16=y^2$

$y = \sqrt{x + 16}$ $y=sqrt(x+16)$

The inverse function should be $f^{- 1} (x) = \sqrt{x + 16}$ $f^-1(x)=sqrt(x+16)$

Answer 2 · 2018-03-21T17:18:15

Please refer to the Explanation.

Suppose that, $f : RR to RR : f(x)=x^2-16$ .

Observe that, $f(1)=1-16=-15, and, f(-1)=-15$ .

$:. f(1)=f(-1)$ .

$:. f" is not injective, or, "1-1$ .

$:. f^-1$ does not exist.

However, if $f$ is defined on a suitable domain, e.g.,

$RR^+$ , then $f^-1$ exists as Respected Serena D. has shown.

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