If f(x)=frac{2x}{4+3|x|}, x in R, then f^{pri | vedclass

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(C) We have, $f(x)=\frac{2x}{4+3|x|}, x \in R$. This can be written as: $f(x) = \begin{cases} \frac{2x}{4+3x}, & 	ext{if } x \geq 0 \ \frac{2x}{4-3x

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$\frac{1}{2}$

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(C) We have, $f(x)=\frac{2x}{4+3|x|}, x \in R$. This can be written as: $f(x) = \begin{cases} \frac{2x}{4+3x}, & 	ext{if } x \geq 0 \ \frac{2x}{4-3x}, & 	ext{if } x To find $f^{\prime}(0)$, we check the left-hand and right-hand derivatives at $x=0$. For $x > 0$, $f^{\prime}(x) = \frac{d}{dx} \left( \frac{2x}{4+3x} ight) = \frac{(4+3x)(2) - 2x(3)}{(4+3x)^2} = \frac{8+6x-6x}{(4+3x)^2} = \frac{8}{(4+3x)^2}$. Thus, $Rf^{\prime}(0) = \lim_{x ightarrow 0^+} \frac{8}{(4+3x)^2} = \frac{8}{16} = \frac{1}{2}$. For $x Thus, $Lf^{\prime}(0) = \lim_{x ightarrow 0^-} \frac{8}{(4-3x)^2} = \frac{8}{16} = \frac{1}{2}$. Since $Lf^{\prime}(0) = Rf^{\prime}(0) = \frac{1}{2}$, the derivative $f^{\prime}(0)$ exists and is equal to $\frac{1}{2}$.

If $f(x)=\frac{2x}{4+3|x|}, x \in R$, then $f^{\prime}(0)=$

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