Need help on both of these Calculus AB mock exams.

AB1

(a) is continuous at if the two-sided limit exists. By definition of , we have . We need to have , since continuity means

By the fundamental theorem of calculus (FTC), we have

We know , and the graph tells us the *signed* area under the curve from 4 to 6 is -3. So we have

and hence is continuous at .

(b) Judging by the graph of , we know has local extrema when . In particular, there is a local maximum when , and from part (a) we know .

For , we see that , meaning is strictly non-decreasing on (6, 10). By the FTC, we find

which is larger than -1, so attains an absolute maximum at the point (10, 7).

(c) Directly substituting into yields

From the graph of , we know the value of the integral is -3 + 7 = 4, so the numerator reduces to 0. In order to apply L'Hopital's rule, we need to have the denominator also reduce to 0. This happens if

Next, applying L'Hopital's rule to the limit gives

which follows from the fact that and .

Then using the value of we found earlier, we end up with

(d) Differentiate both sides of the given differential equation to get

The concavity of the graph of at the point (1, 2) is determined by the sign of the second derivative. For , we have , so the sign is entirely determined by .

When , this is equal to

We know , since has a horizontal tangent at , and from the graph of we also know . So we find , which means is concave upward at the point (1, 2).

(e) is increasing when . The sign of is determined by . We're interested in the interval , and the behavior of on this interval depends on the behavior of on .

From the graph, we know on (0, 4), (6, 9), and (9, 10). This means on (0, 2), (3, 4.5), and (4.5, 5).

(f) Separating variables yields

Integrating both sides gives

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a