Which of the following statements is true for | vedclass

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(B) Given $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \ x^3 & 0 \le x < 1 \ \frac{x^3}{3} - 4x & x < 0 \end{cases}$$1$. Check continuity:At $x = 0$: $f

Vedclass · Accepted Answer

$f'(x)$ fails to exist for $2$ distinct real values of $x$.

Vedclass · Answer

(B) Given $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \ x^3 & 0 \le x $1$. Check continuity:At $x = 0$: $f(0) = 0^3 = 0$. $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (\frac{x^3}{3} - 4x) = 0$. $\lim_{x 	o 0^+} f(x) = 0^3 = 0$. Continuous at $x=0$.At $x = 1$: $f(1) = \sqrt{1} = 1$. $\lim_{x 	o 1^-} f(x) = 1^3 = 1$. $\lim_{x 	o 1^+} f(x) = \sqrt{1} = 1$. Continuous at $x=1$.$2$. Check differentiability:$f'(x) = \begin{cases} \frac{1}{2\sqrt{x}} & x > 1 \ 3x^2 & 0 At $x = 0$: $f'(0^+) = 3(0)^2 = 0$,$f'(0^-) = 0^2 - 4 = -4$. Not differentiable at $x=0$.At $x = 1$: $f'(1^+) = \frac{1}{2\sqrt{1}} = 0.5$,$f'(1^-) = 3(1)^2 = 3$. Not differentiable at $x=1$.$3$. Analyze monotonicity and extrema:For $x  0$ for $x For $0  0$ (increasing).For $x > 1$,$f'(x) = \frac{1}{2\sqrt{x}} > 0$ (increasing).At $x = 0$,$f(0) = 0$. Since $f(x)$ decreases on $(-2, 0)$ and increases on $(0, 1)$,$x = 0$ is a local minimum.$4$. Sign of $f'(x)$:$f'(x) = x^2 - 4$ for $x $f'(x) > 0$ for $x > 0$.Thus,$f'(x)$ changes sign only once at $x = -2$. The statement in option $(C)$ is incorrect as written in the original prompt,but based on the analysis,$f'(x)$ fails to exist at $x=0$ and $x=1$ ($2$ points). Thus,$(B)$ is the correct statement.

Which of the following statements is true for the function $f(x) = \begin{cases} \sqrt{x} & x \ge 1 \\ x^3 & 0 \le x < 1 \\ \frac{x^3}{3} - 4x & x < 0 \end{cases}$

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