$f(x) = \begin{cases} e^x & , \quad 0 \le x \le 1 \ 2 - e^{x-1} & , \quad 1

Question

$f(x) = \begin{cases} e^x & , \quad 0 \le x \le 1 \ 2 - e^{x-1} & , \quad 1 < x \le 2 \ x - e & , \quad 2 < x \le 3 \end{cases}$ and $g(x) = \int_{0}^{x} f(t) dt$ , $x \in [1, 3]$ then

Professor PC Thomas Classes · Answer

By Leibniz's rule, $g'(x) = f(x)$. For extrema, we need $f(x) = 0$. In $(1, 2]$, $2 - e^{x-1} = 0 \implies e^{x-1} = 2 \implies x = 1 + \ln 2$. Here $f(x)$ changes sign from positive to negative, indicating a local maximum at $x = 1 + \ln 2$ (Option C is correct). In $(2, 3]$, $x - e = 0 \implies x = e$. Here $f(x)$ changes sign from negative to positive, indicating a local minimum at $x = e$ (Option D is correct).