If f(x) = text{sgn}(x^3),then | vedclass

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(D) Given $f(x) = 	ext{sgn}(x^3)$.We know that $	ext{sgn}(x^3) = 	ext{sgn}(x)$ because the sign of $x^3$ is the same as the sign of $x$.Thus,$f(x)

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$f$ is not derivable at $x = 0$

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(D) Given $f(x) = 	ext{sgn}(x^3)$.We know that $	ext{sgn}(x^3) = 	ext{sgn}(x)$ because the sign of $x^3$ is the same as the sign of $x$.Thus,$f(x) = \begin{cases} -1, & x  0 \end{cases}$.Since $\lim_{x 	o 0^-} f(x) = -1$ and $\lim_{x 	o 0^+} f(x) = 1$,the function is discontinuous at $x = 0$.$A$ function that is not continuous at a point cannot be derivable at that point.Therefore,$f$ is not derivable at $x = 0$.

If $f(x) = \text{sgn}(x^3)$,then

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