Piecewise Function and Mean Value Theorem. Here's the function:
\begin{align}
f(x) = \left\{ \begin{array}{cc} 
                3 & \hspace{5mm} x=0 \\
                ax+b & \hspace{5mm} 0<x\leq 1 \\
                x^2+3x+c & \hspace{5mm} 1<x\leq 4 \\
                \end{array} \right.
\end{align}
Find the constraints a,b and c so that $f(x)$ satisfies the conditions of MVT.
Here's how I solved. First of all I write the contuinity condition of MVT for $x=1$ and $x=0$
\begin{align}
\lim_{x\to\ 1^+} f(x)=4+c
\end{align}
\begin{align}
\lim_{x\to\ 1^-} f(x)=a+b
\end{align}
\begin{align}
a+b=4+c \hspace{5mm}(1)
\end{align}
And also we can get second limit which only goes one way.
\begin{align}
\lim_{x\to\ 0^+} f(x)=b=f(0) \hspace{5mm} (2)
\end{align}
\begin{align}
b=3
\end{align}
And then I tried to use derivatives for finding other variables. By then, I write teh expressions and I get those:
\begin{align}
\lim_{\Delta x\to\ 0^{+}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=a
\end{align}
\begin{align}
\lim_{\Delta x\to\ 0^{-}} \dfrac{f(1+ \Delta x)-f(1)}{\Delta x}=5
\end{align}
Clearly, $a=5$ and from $(1)$ if we plug all the values that we found we get $c=4$.
Here is my questions:
1-)Is it true?
2-)Are we allowed to do the evaluation that I've done in 2. More explicitly, can I equate just one side to the actual value? Because in this question we don't have any "smaller than 0" part.
Thanks for any help.
 A: A function $f$ is continuous at a point $t$ if $$\lim_{\underset{x \in \text{dom}(f)}{x \to t}} f(x) = f(t)$$
For your function, when $t = 0$, the restriction to the domain of $f$ means we are not looking at any $x < 0$. That is, the condition of continuity becomes
$$\lim_{x \to 0+} f(x) = f(0)$$
Which is exactly the condition you examined in (2).
When $t = 1$, both sides are in the domain, so the condition of continuity is
$$\lim_{x \to 1} f(x) = f(1)$$
But for this piecewise defined function, to examine if this is true, we need to note that $\lim_{x \to 1} f(x)$ exists if and only if the two one-sided limits exist and are equal. That is, if $$\lim_{x \to 1-} f(x) = \lim_{x \to 1+} f(x)$$ which is what you examined in (1). This condition (with (2) and the similar derivative limit) was enough to determine what $a, b, c$ had to be.
But that alone doesn't show that the $f$ defined by those values of $a,b,c$ is actually continuous at $1$. For that, you must still also show that $\lim_{x \to 1} f(x)$, the common value of $\lim_{x \to 1-} f(x)$ and $\lim_{x \to 1+} f(x)$, is in fact $f(1)$ (which follows by the continuity of $x \mapsto ax+b$). I believe the intent of the problem was only to find $a,b,c$, so you don't need to actually make this argument. But all I know is what you wrote here.
