Questions involving $f(x)=\begin{cases}x+1;&0\le x\le1\\2x^2-6x+6,&1\lt x\le2\end{cases}$ and $g(t)=\int_{t-1}^tf(x)dx$ for $t\in[1,2]$

Question

Let $f(x)=\begin{cases}x+1;&0\le x\le1\\2x^2-6x+6,&1\lt x\le2\end{cases}$ and $g(t)=\displaystyle\int_{t-1}^tf(x)dx$ for $t\in[1,2]$. Which of the following hold(s) good?

• A) $f(x)$ is continuous and differentiable in $[0,2]$.
• B) $g'(t)$ vanishes for $t=\frac32$ and $2$
• C) $g(t)$ is maximum at $t=\frac32$
• D) $g(t)$ is minimum at $t=1$

A) $f(x)$ is continuous in $[0,2]$ but non-differentiable at $1$.

B) $g'(t)=f(t)-f(t-1)\implies g'(\frac32)=0=g'(2)$

C) $g''(t)=f'(t)-f'(t-1)\implies g''(\frac32)\lt0$

D) $g'(1)\ne0$, but this option is also given as correct.

Is $g(t)$ really minimum at $t=1$ or could it be a typo?

Answer 1

From the comments,$$g(1)=\displaystyle\int_0^1f(x)dx\\=\displaystyle\int_0^1(x+1)dx\\=\left(\frac{x^2}2+x\right)_0^1=\frac32$$

If we calculate $g(t)$ for $t\in(1,2]$, it would be increasingly greater than $\frac32$. Hence, local minimum is at $x=1$.