Question

Let $ f(x) = \begin{cases} e^x, & 0 \le x \le 1 \\ 2 - e^{x -1}, & 1 < x \le 2 \\ x - e, & 2 < x \le 3 \end{cases} $ and $g(x) = \int\limits^x_0 f(t) dt , x \in [1,3]$ then g(x) has

• A. local maxima at x = 1 + ln 2 and local minima at x = e
• B. local maxima at x = 1 and local minima at x = 2
• C. no local maxima
• D. no local minima

Answer 1

Answer: (A), (B)

$\because g\left(x\right) = \int\limits^{x}_{0} f\left(t\right) dt$
$ \Rightarrow g'\left(x\right) = f\left(x\right) = \begin{cases} e^x, & 0 \le x \le 1 \\ 2 - e^{x -1}, & 1 < x \le 2 \\ x - e, & 2 < x \le 3 \end{cases} $
$\therefore \, g'(x) = 0 \, \Rightarrow \, e^{x -1} = 2 $ or $x -e = 0 $
$\Rightarrow \, x - 1 = \log 2 $ or $x = e$
$\Rightarrow \, x = 1 + ln 2 $ or e
$ g" (x) = \begin{cases} e^x, & 0 \le x \le 1 \\ - e^{x -1}, & 1 < x \le 2 \\ 1, & 2 < x \le 3 \end{cases} $
NOTE THIS STEP
$\therefore \, g" ( 1+ ln 2) = - 2$ and $g" (e) = 1 \, \Rightarrow \, g(x)$ has local max.
at $x = 1 + ln 2$ and local min . at x = e.
Also graph of g(x) suggests, g (x) has local max. at x = 1 and local min. at x = 2