For all real values of x,the increasing funct | vedclass

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(C) function $f(x)$ is strictly increasing if $f'(x) > 0$ for all $x$ in its domain.$(A)$ For $f(x) = x^{-1}$,$f'(x) = -x^{-2} = -\frac{1}{x^2}$,which

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$x^3$

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(C) function $f(x)$ is strictly increasing if $f'(x) > 0$ for all $x$ in its domain.$(A)$ For $f(x) = x^{-1}$,$f'(x) = -x^{-2} = -\frac{1}{x^2}$,which is always negative for $x 
eq 0$.$(B)$ For $f(x) = x^2$,$f'(x) = 2x$,which is negative for $x  0$.$(C)$ For $f(x) = x^3$,$f'(x) = 3x^2$. Since $x^2 \geq 0$ for all real $x$,$f'(x) \geq 0$ for all real $x$. Thus,$f(x) = x^3$ is an increasing function.$(D)$ For $f(x) = x^4$,$f'(x) = 4x^3$,which is negative for $x  0$.Therefore,the correct option is $C$.

For all real values of $x$,the increasing function $f(x)$ is

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