The value of m for which the function f(x) = | vedclass

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(D) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.For continuity,$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) =

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Does not exist

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(D) For $f(x)$ to be differentiable at $x = 1$,it must first be continuous at $x = 1$.For continuity,$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$\lim_{x 	o 1^-} (mx^2) = m(1)^2 = m$.$\lim_{x 	o 1^+} (2x) = 2(1) = 2$.Thus,for continuity,$m = 2$.Now,check for differentiability at $x = 1$ with $m = 2$:Left-hand derivative $L f'(1) = \lim_{h 	o 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h 	o 0} \frac{2(1-h)^2 - 2}{-h} = \lim_{h 	o 0} \frac{2(1 + h^2 - 2h) - 2}{-h} = \lim_{h 	o 0} \frac{2h^2 - 4h}{-h} = \lim_{h 	o 0} (4 - 2h) = 4$.Right-hand derivative $R f'(1) = \lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0} \frac{2(1+h) - 2}{h} = \lim_{h 	o 0} \frac{2h}{h} = 2$.Since $L f'(1) 
eq R f'(1)$ $(4 
eq 2)$,the function is not differentiable at $x = 1$ for any value of $m$.

The value of $m$ for which the function $f(x) = \begin{cases} mx^2, & x \le 1 \\ 2x, & x > 1 \end{cases}$ is differentiable at $x = 1$,is

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