need help on one calculus problem on continuous functions?

-infinity, -4] - not continuous

(-infinity, -4) - not continuous

[-4,-2] - continuous

[-4,-2) - not continuous

[-2,2] - continuous

(-2,2) - not continuous

[2,4] - continuous

[2,4) - not continuous

[4,6] - continuous

(4,6) - not continuous

[6,8] - continuous

(6,8) - not continuous

[8,infinity) - not continuous

(8,infinity) - not continuous

The graph doesn't start at infinity, indicated by the closed dot at 4, it also doesn't end at infinity, indicated by the open dot at 8. For it to have continued to infinity there would have had to have been an arrow pointing into the distance.

[-4,-2]

(-2,2)

[2,4)

(4,6)

(6,8)

Rule out all the intervals with infinity, because the function isn't even defined outside [-4, 8)

Remember that brackets are for closed intervals (which include the endpoints), and parentheses are for open intervals (which exclude the endpoints).

Solid dot at -4, on the continuous curve, up until there's a hole at -2.

That tells us that [-4, -2) is an interval on which g is continuous.

Next we see a dot all by itself at -2.

Then another open dot at -2, on a curve which is continuous up to 2, where there's another hole.

This gives us an interval of (-2, 2)

Then we have a solid dot at 2, on a continuous curve that shoots up toward infinity as it approaches x=4 from the left. So 4 is not included in the interval.

This gives us an interval of [2, 4)

Then there's a curve that shoots up toward infinity as it approaches x=4 from the right. That same curve is continuous until x=6, where there's a hole.

This gives us an interval of (4, 6)

A dot all by itself at x=6 doesn't give us anything.

Then there's another curve which starts from a hole at x=6 and goes until x=8, where there's another hole.

This gives us an interval of (6, 8)

So the whole list of intervals over which the function is continuous:

[-4, -2)

(-2, 2)

[2, 4)

(4, 6)

(6, 8)

you may pick a cost of a to make F non-provide up at 0, yet there is not something which you're able to do to make F non-provide up at useful integer multiples of ?/2. The question is ill conceived. yet so far as 0 is going lim F(x) = lim ax/tan(x) = a x->0+ . . .x->0+ apart from, lim F(x) = lim a² - 2 = a² - 2. x->0- . . ..x->0- F is non-provide up at 0 if those limits agree. this delivers the equation for a a² - 2 = a ==> a² - a - 2 = 0 ==> (a + a million)(a - 2) = 0. F is non-provide up at x = 0 if a = -a million or if a = 2. F can't be made non-provide up at x = ?/2, ?, 3?/2, 2?, and so on.