Subject: Mathematics — Definite Integration
Question: Let $f(x) = \left\{ \begin{matrix}
e^{x}, & 0 \leq x \leq 1 \\
2 - e^{x - 1}, & 1 < x \leq 2 \\
x - e, & 2 < x \leq 3
\end{matrix} \right.\ $ and
$g(x) = \int_{0}^{x}\mspace{2mu} f(t)dt,x \in \lbrack 0,3\rbrack$ then
find points of maxima
and minima of $g(x)$
Answer: $g(x)$ has local maxima at $x = 1 + ln2,\ 1$ and local
minima at $x = e,2$
Solution: Given
$g(x) = \int_{0}^{x}\mspace{2mu} f(t)dt,x \in \lbrack 0,3\rbrack$\
$${\Rightarrow g^{'}(x) = f(x) = \left\{ \begin{matrix}
e^{x}, & 0 \leq x \leq 1 \\
2 - e^{x - 1}, & 1 < x \leq 2 \\
x - e, & 2 < x \leq 3
\end{matrix} \right.\ 
}{\therefore g^{'}(x) = 0 \Rightarrow e^{x - 1} = 2\text{~or~}x - e = 0
}{\Rightarrow x - 1) = ln2\text{~or~}x = e \Rightarrow x = 1 + ln2\text{~or~}e
}{g^{''}(x) = \left\{ \begin{matrix}
e^{x}, & 0 \leq x \leq 1 \\
 - e^{x - 1}, & 1 < x \leq 2 \\
1, & 2 < x \leq 3
\end{matrix} \right.\ 
}$$$g(x)$ has local max, at $x = 1 + ln2$ and local min. at $x = e$.
[IMAGE] Also from graph of $g^{'}(x)$, it is clear that $g(x)$ has local max. at
$x = 1$ and local min. at $x = 2$