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(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
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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
