Let
$$f(x)=\{\begin{array}{ll}0& \text{ if }x<-5\\ 3& \text{ if }-5\le x<0\\ -2& \text{ if }0\le x<3\\ 0& \text{ if }x\ge 3\end{array}$$
and
$$g(x)={\int }_{-5}^{x}f(t)dt$$
Determine the value of each of the following:
(a) $g(-7)=$
(b) $g(-4)=$
(c) $g(1)=$
(d) $g(4)=$
(e) The absolute maximum of $g(x)$ occurs when $x=$ and is the value
It may be helpful to make a graph of $f(x)$ when answering these questions.

Step 1 ：The function $g(x)$ is the integral of $f(x)$ from $-5$ to $x$. This means that $g(x)$ is the area under the curve of $f(x)$ from $-5$ to $x$.

Step 2 ：To find the value of $g(x)$ at a specific point, we need to calculate the area under the curve of $f(x)$ from $-5$ to that point.

Step 3 ：For $g(-7)$, since $-7<-5$, the integral from $-5$ to $-7$ does not exist, so $g(-7)=0$.

Step 4 ：For $g(-4)$, we need to calculate the area under the curve from $-5$ to $-4$. Since $f(x)=3$ for $-5\le x<0$, the area is a rectangle with height 3 and width 1, so $g(-4)=3\ast 1=3$.

Step 5 ：For $g(1)$, we need to calculate the area under the curve from $-5$ to $1$. This is the sum of the area from $-5$ to $0$ and the area from $0$ to $1$. The area from $-5$ to $0$ is a rectangle with height 3 and width 5, so it is $3\ast 5=15$. The area from $0$ to $1$ is a rectangle with height $-2$ and width 1, so it is $-2\ast 1=-2$. Therefore, $g(1)=15-2=13$.

Step 6 ：For $g(4)$, we need to calculate the area under the curve from $-5$ to $4$. This is the sum of the area from $-5$ to $0$, the area from $0$ to $3$, and the area from $3$ to $4$. The area from $-5$ to $0$ is $3\ast 5=15$, the area from $0$ to $3$ is $-2\ast 3=-6$, and the area from $3$ to $4$ is $0\ast 1=0$. Therefore, $g(4)=15-6+0=9$.

Step 7 ：The absolute maximum of $g(x)$ occurs when the area under the curve is maximized. This happens at $x=0$, where the area under the curve from $-5$ to $0$ is maximized and is the value $15$.

Step 8 ：Final Answer: (a) $g(-7)=\boxed{{\displaystyle 0}}$ (b) $g(-4)=\boxed{{\displaystyle 3}}$ (c) $g(1)=\boxed{{\displaystyle 13}}$ (d) $g(4)=\boxed{{\displaystyle 9}}$ (e) The absolute maximum of $g(x)$ occurs when $x=\boxed{{\displaystyle 0}}$ and is the value $\boxed{{\displaystyle 15}}$.