# Let f(x)= 0 if x is less than -5 f(x)= 2 if x is greater than or equal to -5 but less than 1f(x)= -4 if x is greater than or equal to 1 but less than 4 f(x)= 0 if x is greater than or equal to 4...

Let

f(x)= 0 if x is less than -5

f(x)= 2 if x is greater than or equal to -5 but less than 1

f(x)= -4 if x is greater than or equal to 1 but less than 4

f(x)= 0 if x is greater than or equal to 4

and g(x)= integrate from -5 to x of (f(t)dt)

Determine the value of each of the following:

a) The absolute maximum of g(x) occurs when x= ? and is the value ?

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First, let's look at a graph of f:

When f is 0, then g is staying constant.

So g is constant when x<-5

When f is positive, then g is increasing. So g is increasing from x=-5 to x = 1.

When f is negative, then g is decreasing. So g is decreasing from x=1 to x=4.

And again, g is constant when x>4

Thus the maximum of g must occur at x=1.

`g(1)=int_(-5)^1 f(t) dt`

We can use calculus, but there is something easier:

An integral is just the area under the curve. Thus we want the area under f from x=-5 to x=1

f has a height of 2 there, and the width from -5 to 1 is 6. Thus the area is just the area of a rectangle: 2*6=12

So the maximum for g occurs at x=1, and there g is 12.