Mathematics - Application of Derivative Question with Solution | TestHub

Here, $f(x)=\{\begin{array}{cc}{e}^x,& 0\le x\le 1\\ 2-e^{x-1},& 1<x\le 2\\ x-e,& 2<x\le 3\end{array}$ and $g(x)={\int }_0^{x}f(t)dt$ $$\Rightarrow g^{\prime }(x)=f(x)$$ where, $g^{\prime }(x)=0$ $\Rightarrow x=1+{\log }_{e}2$ and $x=e$. Also, $g(x)=\{\begin{array}{cc}-e^{x-1},& 1<x\le 2\\ 1,& 2<x\le 3\end{array}$ $g\left(1+{\log }_{e}2\right)=-e^{{\log }_{e}2}<0$, hence at $x=1+{\log }_{e}2,g(x)$ has a local maximum. Also, $g(e)=1>0$, hence at $x=e,g(x)$ has a local minima. $\because f(x)$ is discontinuous at $x=1$, then we get Local maxima at $x=1$ and local minima at $x=2$.