Question

Let $f(x)=\{\begin{array}{ll}{e}^x,& 0\le x\le 1\\ 2-e^{x-1},& 1<x\le 2\\ x-e,& 2<x\le 3\end{array}$ and $g(x)=\underset{0}{\overset{x}{\int }}f(t)dt,x\in [1,3]$ 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)

$\because g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
$\Rightarrow g^{\prime }\left(x\right)=f\left(x\right)=\{\begin{array}{ll}{e}^x,& 0\le x\le 1\\ 2-e^{x-1},& 1<x\le 2\\ x-e,& 2<x\le 3\end{array}$
$\therefore g^{\prime }(x)=0\Rightarrow e^{x-1}=2$ or $x-e=0$
$\Rightarrow x-1=\log 2$ or $x=e$
$\Rightarrow x=1+ln2$ or e
$g"(x)=\{\begin{array}{ll}{e}^x,& 0\le x\le 1\\ -e^{x-1},& 1<x\le 2\\ 1,& 2<x\le 3\end{array}$
NOTE THIS STEP
$\therefore g"(1+ln2)=-2$ and $g"(e)=1\Rightarrow g(x)$ has local max.
at $x=1+ln2$ 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