1. ## Equalling Derivatives

Suppose the function g is defined by
g(x) = k $sqrt(x + 1)$ for 0 [LaTeX ERROR: Compile failed] x [LaTeX ERROR: Compile failed] 3

= mx + 2 for 3 < x [LaTeX ERROR: Compile failed] 5

where k and m are constants. If g is differentiable at x = 3, what are the values of k and m.
this is part c on a question, but i dont think you need the information from the original question.
Thanks!

2. If it is given that $g(x)$ is differentiable at $x=3$, then both pieces of the piecewise function must have the same values of $g(3)$ and $g'(3)$.

So let's see where the two pieces are equal to each other when $x=3$:

$k\sqrt{3+1}=3m+2$
$k=\frac{3m+2}{2}$

Making that substitution, let's see where the derivatives are equal to each other at $x=3$:

$\frac{\frac{3m+2}{2}}{2\sqrt{3+1}}=m$
$\frac{3m+2}{8}=m$
$m=\frac{2}{5}$
$k=\frac{3(\frac{2}{5})+2}{2}=\frac{8}{5}$

So, $g(x)$ is differentiable at $k=\frac{8}{5},m=\frac{2}{5}$. You can check your work to see that this piecewise function is continuous at $x=3$ with these constants and that there is a single value for $g'(3)$ with these constants.

3. Originally Posted by Pinkk
If it is given that $g(x)$ is differentiable at $x=3$, then both pieces of the piecewise function must have the same values of $g(3)$ and $g'(3)$.

So let's see where the two pieces are equal to each other when $x=3$:

$k\sqrt{3+1}=3m+2$
$k=\frac{3m+2}{2}$

Making that substitution, let's see where the derivatives are equal to each other at $x=3$:

$\frac{\frac{3m+2}{2}}{2\sqrt{3+1}}=m$
$\frac{3m+2}{8}=m$
$m=\frac{2}{5}$
$k=\frac{3(\frac{2}{5})+2}{2}=\frac{8}{5}$

So, $g(x)$ is differentiable at $k=\frac{8}{5},m=\frac{2}{5}$. You can check your work to see that this piecewise function is continuous at $x=3$ with these constants and that there is a single value for $g'(3)$ with these constants.
thanks so much. its so simple once u explain it