QuestionSuppose $f(x)=8^{x}$ . a) What is the domain of $f^{-1}(x)$ ? Enter your answer using interval notation. 1 [] U $\infty$ $\pi$ $a^{\circ}$ $a^{b}$ $\sqrt{a}$ $\square$ $\sin$ $\square$ Previev Invalid Inpu b) What is the range of $f^{-1}(x)$ ? Enter your answer using interval notation.

Studdy Solution

STEP 1

1. The function $f(x) = 8^x$ is an exponential function.
2. The inverse function $f^{-1}(x)$ will reverse the roles of the input and output of $f(x)$ .
3. The domain of a function is the set of all possible input values.
4. The range of a function is the set of all possible output values.
5. The domain of $f(x)$ is $(-\infty, \infty)$ because an exponential function is defined for all real numbers.
6. The range of $f(x)$ is $(0, \infty)$ because an exponential function with a positive base never equals zero or a negative number.

STEP 2

1. Determine the domain of $f^{-1}(x)$ .
2. Determine the range of $f^{-1}(x)$ .

STEP 3

Recall that the domain of the inverse function $f^{-1}(x)$ is the range of the original function $f(x)$ .
Since $f(x) = 8^x$ , the range of $f(x)$ is $(0, \infty)$ .
Therefore, the domain of $f^{-1}(x)$ is $(0, \infty)$ .
$\text{Domain of } f^{-1}(x) = (0, \infty)$

STEP 4

Recall that the range of the inverse function $f^{-1}(x)$ is the domain of the original function $f(x)$ .
Since the domain of $f(x)$ is $(-\infty, \infty)$ ,
the range of $f^{-1}(x)$ is $(-\infty, \infty)$ .
$\text{Range of } f^{-1}(x) = (-\infty, \infty)$
Solution: a) The domain of $f^{-1}(x)$ is $(0, \infty)$ . b) The range of $f^{-1}(x)$ is $(-\infty, \infty)$ .

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