Our function of interest is defined by an accumulation

\begin{equation*}
f(x) = 10 + \int_0^x f'(z) \, dz\text{.}
\end{equation*}

We can find local extrema in the same way as before. However, instead of solving an equation \(f'(x)=0\text{,}\) we can look at the graph to both find the roots and the signs of \(f'\text{.}\) The graph crosses the \(x\)-axis at \(x=-3\text{,}\) \(x=1\) and \(x=5\text{.}\) The signs of \(f'(x)\) are identified in the following sign analysis number line summary.

The Theorem 6.2.3 allows us to conclude that \(f(x)\) has a local maximum at \(x=-3\text{,}\) a local minimum at \(x=1\text{,}\) and another local maximum at \(x=4\text{.}\) It is possible to decide which maximum has a higher value by considering the signed area of the graph. In particular, because \(f'(x)\) has linear segments, we can compute the areas in question using elementary geometry to find

\begin{gather*}
\int_{-3}^{1} f'(x) \, dx = -5,\\
\int_{1}^{4} f'(x) \, dx = 6.
\end{gather*}

Using the splitting property of definite integrals, this implies

\begin{equation*}
f(4)-f(-3) = \int_{-3}^{4} f'(x) \, dx = -5 + 6 = 1\text{.}
\end{equation*}

Consequently, \(f(4) = f(-3) + 1\) and \(f\) has a higher value at \(x=4\) than at \(x=-3\text{.}\)

We can find actual values if we use the initial value,

\begin{equation*}
f(-3) = 10 + \int_{0}^{-3} f'(z) \, dz\text{.}
\end{equation*}

Because the integral goes right to left, we have \(dz \lt 0\) and the signed area will be negated. Again using geometry, we find

\begin{equation*}
f(-3) = 10 + -(-4.5) = 14.5 \text{.}
\end{equation*}

Using this point and the integrals above, we can quickly find

\begin{align*}
f(1) &= f(-3) + \int_{-3}^{1} f'(z) \, dz = 14.5 -5 = 9.5,\\
f(4) &= f(1) + \int_{1}^{4} f'(z) \, dz = 9.5 +6 = 15.5.
\end{align*}

To find global extrema, we need to think about what happens to the left and right of these local extrema. The sign analysis of \(f'(x)\) shows that \(f\) is increasing on \((-\infty,-3)\text{.}\) Thus, \(f(-3)\) is the maximum value on \((-\infty,3]\text{.}\) The value of \(f\) is unbounded below on this interval,

\begin{equation*}
\lim_{x \to -\infty} f(x) = -\infty\text{.}
\end{equation*}

We can see this by considering the integral \(\displaystyle \int_{-3}^{x}\) for \(x \lt -3\text{.}\) Because the integral goes right to left, \(dz \lt 0\text{,}\) and \(f'(z) \gt 0\) on this interval, the integral becomes more and more negative the further \(x\) goes to the left. With a similar argument using \(\displaystyle \int_4^x f'(z) \, dz\text{,}\) we find

\begin{equation*}
\lim_{x \to \infty} f(x) = -\infty
\end{equation*}

and \(f\) is again unbounded below on the interval \((4,\infty)\text{.}\)

We conclude that \(f\) has a global maximum \(f(4) = 13.5\) and no global minimum. A graph of \(y=f(x)\) is shown below.