Q:

Prove that $g'(t)$ vanishes for $t=\frac{3}{2}$ and $2,g(t)$ is maximum when $t=\frac{3}{2}$ and $g(t)$ is minimum at $t=1.$

Let $f(x)= \left\{ \begin{array}{lcc}
             x+1,  & 0 \leq x \leq 1 \\
             \\ 2x^2-6x+6, & 1 < x \leq 2 \\
             \\ 
             \end{array}
   \right.$ and $g(t)=\int_{t-1}^{t}f(x)dx$ for $t\in[1,2]$
Then prove that $g'(t)$ vanishes for $t=\frac{3}{2}$ and $2,g(t)$ is maximum when $t=\frac{3}{2}$ and $g(t)$ is minimum at $t=1.$

I found $g'(t)=f(t)-f(t-1)$
Put $g'(t)=f(t)-f(t-1)=0$
As $t\in[1,2]$ so $f(t)=2t^2-6t+6$ and as $t-1\in[0,1]$ so $f(t-1)=(t-1)+1=t$
$g'(t)=2t^2-7t+6=0$ gives $t=\frac{3}{2}$ and $2$.
$g''(t)=4t-7$
$g''(\frac{3}{2})=-1$,so $g(t)$ is maximum when $t=\frac{3}{2}$
But i do not understand how $g(t)$ is minimum at $t=1.$
Please help me.

A:

You are looking for absolute maximum of a continuous function $g(t)$ in a closed interval. 
You found two critical points $\frac{3}{2}$ and $2$.
The next step is to test the values of $g(t)$ at these critical points, plus the boundary points $1$ and $2$.
After comparing their values as the following, you can find the maximum and minimum:
$$g\left(\frac{3}{2}\right)=\int^{\frac{3}{2}}_{\frac{1}{2}} f(x)\, dx\\
g\left(2\right)=\int^{2}_{1} f(x)\, dx\\
g\left(1\right)=\int^{1}_{0} f(x)\, dx$$

