Question

g(t)={6-pt ,t<0 {q+e^(-2t) ,t<=0 is differentiable at t=0,what is the value of g(q-p)

Answer 1

\[g(t)=\begin{cases}6-pt&\text{for }t\color{red}<0\\q+e^{-2t}&\text{for }t\color{red}{\le}0\end{cases}\]
Please check those inequality signs again.

You're told that \(g(t)\) is differentiable at \(t=0\), which means the derivative of \(g\) exists.

For the derivative to exist at a point \(t\), you must have that \(g(t)\) exists and \(\displaystyle \lim_{t\to c^-}g'(t)=\lim_{t\to c^+}g'(t)\) (which basically means the slope of the tangent line at \(t\) doesn't suddenly change).

Answer 2

Here's what you have
\[g'(t)=\begin{cases}-p&\text{for }t<0\\?&\text{for }t=0\\-2e^{-2t}&\text{for }t>0\end{cases}\]
(Again, please check the signs there. I'm guessing.)
We know that \(g'(0)\) exists and allows for continuity of the derivative, so you know that \(\displaystyle\lim_{t\to0}g'(t)=g'(0)\).

For this to be true, you must have that the one-sided limits exist (I mentioned this in my first post):
\[\begin{align*}\lim_{t\to 0^-}g'(t)&=\lim_{t\to 0^+}g'(t)\\
\lim_{t\to0^-}(-p)&=\lim_{t\to0^+}-2e^{-2t}\\
-p&=-2\\
p&=2
\end{align*}\]
So, though it's not important to know, you have that \(g'(0)=2\).

Answer 3

What remains is to solve for \(q\). We know the derivative exists at 0, which means \(g\) is continuous at \(t=0\).

This means the one-sided limits as \(t\to0\) exist and are the same from both sides:
\[\begin{align*}\lim_{t\to 0^-}g(t)&=\lim_{t\to 0^+}g(t)\\
\lim_{t\to0^-}(6-2t)&=\lim_{t\to0^+}\left(q+e^{-2t}\right)
\end{align*}\]
I'll let you do the rest. Solve for \(q\), then find \(g(q-p)\).