If f(x) = begin{cases} frac{8}{x^3} - 6x, & x | vedclass

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(B) To check continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = \frac{8}{(1)^3} - 6(1) = 8 - 6 = 2$. $\lim_{x 	o 1^+} f(x) = \sqrt{1} + 1 = 1 + 1 = 2$.

Vedclass · Accepted Answer

continuous but not differentiable

Vedclass · Answer

(B) To check continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = \frac{8}{(1)^3} - 6(1) = 8 - 6 = 2$. $\lim_{x 	o 1^+} f(x) = \sqrt{1} + 1 = 1 + 1 = 2$. Since $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1) = 2$, the function is continuous at $x = 1$. To check differentiability at $x = 1$: $	ext{LHD} = \frac{d}{dx} (\frac{8}{x^3} - 6x) = -24x^{-4} - 6$. At $x = 1$, $	ext{LHD} = -24 - 6 = -30$. $	ext{RHD} = \frac{d}{dx} (\sqrt{x} + 1) = \frac{1}{2\sqrt{x}}$. At $x = 1$, $	ext{RHD} = \frac{1}{2}$. Since $	ext{LHD} 
eq 	ext{RHD}$, the function is not differentiable at $x = 1$. Therefore, $f$ is continuous but not differentiable at $x = 1$.

Column $I$	Column $II$
$(A)$ $f(x) = x\|x\|$	$(p)$ continuous in $(-1, 1)$
$(B)$ $f(x) = \sqrt{\|x\|}$	$(q)$ differentiable in $(-1, 1)$
$(C)$ $f(x) = x + [x]$	$(r)$ strictly increasing in $(-1, 1)$
$(D)$ $f(x) = \|x - 1\| + \|x + 1\|$	$(s)$ not differentiable at least at one point in $(-1, 1)$

If $f(x) = \begin{cases} \frac{8}{x^3} - 6x, & x \le 1 \\ \sqrt{x} + 1, & x > 1 \end{cases}$, then at $x = 1$, $f$ is:

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