Consider the following for the two (02) items that follow: Let f(x)...

The given function can be rewritten as:

$f(x) =$\begin{cases} x^2, & x \leq -1 \\ x^3, & -1 < x < 1 \\ x^2, & x \geq 1 \end{cases}$$<br><br>Checking continuity at$x = -1$:<br>Left Hand Limit (LHL)$= \lim_{x \to -1^-} x^2 = (-1)^2 = 1$<br>Right Hand Limit (RHL)$= \lim_{x \to -1^+} x^3 = (-1)^3 = -1$<br>Since LHL ≠ RHL,$f(x)$is not continuous at$x = -1$. Statement I is incorrect.<br><br>Checking differentiability at$x = 1$:<br>First, check continuity at$x = 1$:<br>LHL$= \lim_{x \to 1^-} x^3 = 1$<br>RHL$= \lim_{x \to 1^+} x^2 = 1$<br>$f(1) = 1^2 = 1$<br>The function is continuous at$x = 1$.<br><br>Now, check the derivatives at$x = 1$:<br>Left Hand Derivative (LHD)$= \lim_{x \to 1^-} \dfrac{d}{dx}(x^3) = \lim_{x \to 1^-} 3x^2 = 3(1)^2 = 3$<br>Right Hand Derivative (RHD)$= \lim_{x \to 1^+} \dfrac{d}{dx}(x^2) = \lim_{x \to 1^+} 2x = 2(1) = 2$<br>Since LHD ≠ RHD,$f(x)$is not differentiable at$x = 1$. Statement II is incorrect.

Both statements are incorrect.

Answer: Neither I nor II