a)
We have the following function given by an integral

Let's evaluate it at g(-1), we have

Looking at the graph of f(t) we can see that it's the area of a triangle of base 1 and height 3, therefore

g(-1) is -3/2

To find the derivative of f(x) we can use the fundamental theorem, therefore

Therefore

Looking at the graph we can see that, g'(-1) = 0

And now for the last derivative, we have

As we can see f(x) is a line of positive slope at the interval (-∞, 0], the derivative of a line if the slope, we can easily see that the slope of f(x) at (-∞, 0] is 3, therefore

d) Here we will use the fact that f is compounded function (or a modular function), so let's find the expression of f and then do the integral to find g, as we can see, we have a line for (-∞, 0] and another line for [0, ∞), the equations are very easy to find.
For (-∞, 0] we have

And for [0, ∞)

Then we can write the function f

Therefore when we solve the integral for g(x) we will find two different functions, one for x>0 and the other for x<0. then let's do it first for x<0

Now we will repeat the exact same process but now for x > 0

Therefore we have the expression for g now

Then to graph g we must graph two quadratics! Let's plot these quadratics from -2 to 2