Answer
$a=3, m=1, b=4$
Work Step by Step
The Mean Value Theorem requires
- that $f(x)$ is continuous on $[a,b]$
- that $f'(x)$ is defined on $(a,b)$
To be continuous, f must be continuous at the endpoints 0 and 2, as well as at x=1.
At x=0, f needs to be right-continuous
$\displaystyle \lim_{x\rightarrow 0^{+}}f(x)=a$, and this needs to be equal to $f(0)=3$
$\Rightarrow a=3$
At $x=1$ both one sided limits must equal $f(1)=m(1)+b=m+b$
$\displaystyle \lim_{x\rightarrow 1^{-}}f(x)=-1+3+3=5$,
So, $m+b=5\qquad (*)$
$f'(x)=\left\{\begin{array}{ll}
0, & x=0\\
-2x+3, & 0\lt x\lt 1\\
m & 1\leq x\leq 0
\end{array}\right.$
For $f'(x)$ to be defined at $x=1, $ we must have
$\displaystyle \lim_{x\rightarrow 1^{-}}f('x)=\lim_{x\rightarrow 1^{+}}f(x)$
$-2+3=m$
$m=1$
Back substituting into $(*),\quad b=4$
