If the function g(x)=caseskx+1, & 0 x3+2, & 3

Question

If the function $g(x)=\begin{cases}k\sqrt{x+1}, & 0\le x\le3\mx+2, & 3<x\le5\end{cases}$ is differentiable, then the value of $k+m$ is:

Accepted Answer

Continuity at $x=3$: $2k=3m+2$ ... (1)

Differentiability at $x=3$: $\dfrac{k}{2\sqrt3+1}$... more precisely $\dfrac{k}{2\cdot2}=m\Rightarrow k=4m$ ... (2)

Solving: $m=\dfrac25,\ k=\dfrac85$. So $k+m=2$.

Solutions · 1