What is the range of the inverse function of $f(x)=$ $x^{3}$ ?
$f^{-1}(x)> 0$
$f^{-1}(x)< 0$
$f^{-1}(x) \geq 0$
$f^{-1}(x) \leq 0$
All real numbers

Answer

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Answer

$\boxed{\text{The range of } f^{-1}(x) = \sqrt[3]{x} \text{ is all real numbers.}}$

Steps

Step 1 ：The inverse function of $f(x) = x^{3}$ is $f^{-1}(x) = \sqrt[3]{x}$.

Step 2 ：The cube root function, $\sqrt[3]{x}$, is defined for all real numbers. This is because you can take the cube root of any number, positive, negative, or zero.

Step 3 ：Therefore, the range of the inverse function $f^{-1}(x) = \sqrt[3]{x}$ is all real numbers.

Step 4 ：$\boxed{\text{The range of } f^{-1}(x) = \sqrt[3]{x} \text{ is all real numbers.}}$

What is the range of the inverse function of $f(x)=$ $x^{3}$ ?$f^{-1}(x)> 0$$f^{-1}(x)< 0$$f^{-1}(x) \geq 0$$f^{-1}(x) \leq 0$All real numbers

Answer

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What is the range of the inverse function of $f(x)=$ $x^{3}$ ?
$f^{-1}(x)> 0$
$f^{-1}(x)< 0$
$f^{-1}(x) \geq 0$
$f^{-1}(x) \leq 0$
All real numbers