This function is continuous because the limit on the left and the limit on the right are equal to the value of the function at \(x=2\text{,}\) as follows:

\begin{align*}
\lim_{x \to 2^-}f(x) &= \lim_{x \to 2}(x^2-3x+8) = 2^2-3(2)+8 = 6,\\
\lim_{x \to 2^+}f(x) &= \lim_{x \to 2}(5x-x^2) = 5(2)-2^2 = 6,\\
f(2) &= 5(2)-2^2 = 6.
\end{align*}

Now that we know the function is continuous, we can compute the derivative using left- and right-limits. Because \(f(x) = x^2-3x+8\) for \(x \lt 2\text{,}\) we have

\begin{equation*}
f(2+h) = (2+h)^2-3(2+h)+8 = 6+h+h^2
\end{equation*}

when \(h \lt 0\text{.}\) Consequently, the derivative from the left is

\begin{align*}
\frac{df}{dx}\Big|_{2^-} &= \lim_{h \to 0^-} \frac{f(2+h)-f(2)}{h}\\
&= \lim_{h \to 0^-} \frac{(6+h+h^2) - 6}{h} \\
&= \lim_{h \to 0^-} \frac{h(1+h)}{h} \\
&= \lim_{h \to 0^-} 1+h \\
&= 1+0 = 1
\end{align*}

When \(h \gt 0\text{,}\) we have

\begin{equation*}
f(2+h) = 5(2+h)-(2+h)^2 = 6+h-h^2 \text{.}
\end{equation*}

The derivative from the right is

\begin{align*}
\frac{df}{dx}\Big|_{2^+} &= \lim_{h \to 0^+} \frac{f(2+h)-f(2)}{h}\\
&= \lim_{h \to 0^+} \frac{(6+h-h^2) - 6}{h} \\
&= \lim_{h \to 0^+} \frac{h(1-h)}{h} \\
&= \lim_{h \to 0^+} 1-h \\
&= 1-0 = 1
\end{align*}

Since the left and right limits computing the left and right derivatives are the same, we conclude that

\begin{equation*}
\frac{df}{dx}(2) = \lim_{h \to 0}\frac{f(2+h)-f(2)}{h} = 1
\end{equation*}

So \(f\) is differentiable at \(x=2\text{.}\)

The function consists of two parabolas joined together at \(x=2\text{.}\) When the left and right derivatives agree, the function transitions smoothly with no corner at \(x=2\text{.}\)