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 $x = 1$.

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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