[tex]\bf f(x)=
\begin{cases}
4,&x=0\\
ax+b,&0\ \textless \ x\le 1\\
x^2+5x+c,&1\ \textless \ x\le 2
\end{cases}[/tex]
the mean value theorem states, that the function for such value to exist, must be continuous and differentiable, graph wise, that means the graph of the function between those bounds, must be "a smooth line", without any abrupt jumps or changes like a spike or a sharp sink.
so, rewording the material some, the piece-wise function has 3 subfunctions, and within the interval of [0, 2], those 3 must meet seamlessly at the junction points, namely 0 and 1, so the "4" part must seamlessly blend in with the ax + b part at 0, and the ax + b part must meet seamlessly with the x²+5x+c.
that said, we could simply freely pick a "c" value for the third subfunction, and from there, make the second subfunction to meet it at that point, so let's do so.
let's say, we pick c = 3, so the 3rd subfunction is x² + 5x + 3.
now, what's f(x) for that parabola at x = 1?, well, is 1²+5(1)+3, or 9, so it lands on the ( 1, 9 ) point then.
now, the "4" subfunction is just a horizontal line, well, at x = 0, is really just a dot, so, ax + b must meet that dot and also must meet the parabola at (1, 9), given that c = 3.
so, now, what's the equation of a line that passes through (0, 4) and (1, 9)?
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 0}}\quad ,&{{ 4}})\quad
% (c,d)
&({{ 1}}\quad ,&{{ 9}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{9-4}{1-0}\implies 5
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-4=5(x-0)\implies \boxed{y=5x+4}[/tex]
so, that'd be the 2nd subfunction, meeting "4" and x² + 5x + 3, and surely you can see what "a" and "b" are.
check the picture below.
and since they meet seamlessly, they meet the mean value theorem at [0, 2].
