Question

Determine values of k and m such that the following function g(x) is continuous and differentiable at all points.
g(x) = 
3x + k if x < 0
e
mx if x ≥ 0

Answer 1

g(x) = 3x + k if x < 0
emx if x ≥ 0

Answer 2

i think i have the first part but i'm wondering how to show it differtiable at all points! any idea?

Answer 3

\[g(x)=\begin{cases}3x+k&\text{for }x<0\\
emx&\text{for }x\ge0\end{cases}\]
Does \(emx\) use \(e\) as in \(2.718...\)? Maybe you mean \(e^{mx}\)?

Whatever it may be, for \(g\) to be continuous at 0, you must have the following satisfied:
\[\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)=g(0)\]
For \(g\) to be differentiable at 0, you must have
\[\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\]
If these two limits are the same, then they are also equal to \(g'(0)\). To find the corresponding \(g'(x)\)'s, you would differentiate one piece of the function at a time:
\[g(x)=\begin{cases}3x+k&\text{for }x<0\\
emx&\text{for }x\ge0\end{cases}~~\Rightarrow~~g'(x)=\begin{cases}\dfrac{d}{dx}(3x+k)&\text{for }x<0\\\\
?&\text{for }x=0\\
\dfrac{d}{dx}(emx)&\text{for }x>0\end{cases}\]
The question mark indicates that we don't know what \(g'(0)\) is, or if it even exists. It exists if the one-sided derivative limits exist and are the same. It does not exist if either the limits do not exist or are not the same.

Answer 4

Sorry for the late reply. You are correct when you say e^mx! Thank You for the help! i think I know what you mean now.