If the function g(x)=begin{cases} K sqrt{x+1} | vedclass

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(B) For the function $g(x)$ to be differentiable at $x=3$, it must first be continuous at $x=3$. Continuity at $x=3$: $\lim_{x 	o 3^-} g(x) = \lim_{x

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$2$

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(B) For the function $g(x)$ to be differentiable at $x=3$, it must first be continuous at $x=3$. Continuity at $x=3$: $\lim_{x 	o 3^-} g(x) = \lim_{x 	o 3^+} g(x) = g(3)$. $K \sqrt{3+1} = 3m+2 \implies 2K = 3m+2$ (Equation $1$). Differentiability at $x=3$: The left-hand derivative must equal the right-hand derivative. $g'(x) = \begin{cases} \frac{K}{2\sqrt{x+1}} &, 0 At $x=3$, $\frac{K}{2\sqrt{3+1}} = m \implies \frac{K}{4} = m \implies K = 4m$ (Equation $2$). Substitute $K=4m$ into Equation $1$: $2(4m) = 3m+2 \implies 8m = 3m+2 \implies 5m = 2 \implies m = \frac{2}{5}$. Then $K = 4(\frac{2}{5}) = \frac{8}{5}$. Therefore, $K+m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2$.

If the function $g(x)=\begin{cases} K \sqrt{x+1} &, 0 \leq x \leq 3 \\ mx+2 &, 3 < x \leq 5 \end{cases}$ is differentiable, then $K+m=$

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