It's calculus work. the screenshot below should answer my question. 80 points

a) In order for to be continuous at , we need to have

By definition of , we know . Meanwhile, the limits are

so is indeed continuous at .

b) We use the first derivative test (FDT) here, but when we compute the derivative of a piecewise function, we have to be careful at the points where the pieces "split off", because it's possible that the derivative does not exist at these points, yet an extreme value can still occur there. (Consider, for example, at .)

In this case,

We find the critical points for each piece over their respective domains:

On the first piece:

which does fall in [-2, 2]. The FDT shows for less than and near -1, and for greater than and near -1, so is a local maximum.

On the second piece:

so it does not contribute any critical points.

Where the pieces meet:

By checking the conditions for continuity mentioned in part (a), we can determine that does not exist, but that doesn't rule out as a potential critical point.

We have

so for less than and near 1, and

so for greater than and near 1. So the FDT tells us that is a local minimum.

Finally, at the endpoints of the domain we're concerned with, [-2, 2]:

We have and .

So, on [-2, 2], attains an absolute minimum of and an absolute maximum of .

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph