The exam is here.
Problem 1 asks us to compute
Since direct substitution is not possible (leading to ), we try a different approach.
Method 1: We use L’Hôpital’s rule. Recall that one of the versions of the rule states that if:
- , and
- , then
(There is another version of the rule, where in item 1. we have instead that and , but we do not need this version here.)
Since and , and
it follows that
Method 2: We use directly the definition of derivative. Recall that for any function and any ,
In particular, letting and , we have so and
(We have used instead of , but of course the name of a variable does not matter.)
Similarly, in lecture we saw that
(This is another instance of the same idea: Letting , we have and
but the fact that ultimately traces back to (see Section 4.3 in the textbook), so it is better to simply use this limit directly.)
We therefore have
(Naturally, this coincides with the answer obtained by the first method.)
Problem 2 asks to find if is implicitly given by the equation
We differentiate both sides of this identity, using the product rule and the chain rule in the computation of the left-hand side. We get:
We now solve this new equation for : or , or
In this particular equation, this method is essentially the only way to compute , since we cannot solve the equation directly and find a closed formula for as a function of .
The graphs below show the set of points with . Actually, plotting the graph is (computationally) harder than one would have anticipated. The first one is a poor rendering of the actual curve, and it was obtained with Maple 12 (there are Maple commands to improve the level of detail and get better versions, though). The second one is a bit better, and it was obtained with Grapher (but one still sees a few features that are mistakes introduced by the algorithms used to produce graph).
By the way, note how the graph is most definitely not the graph of a function. For example, for the values of for which there is at least one satisfying , in fact there are at least two (except when ). This can be quickly checked: If and then as well. Moreover, for any , satisfies . In order to make sense of , we must therefore restrict ourselves to an appropriate domain and range. As it is the case with the inverse trigonometric functions, we have several options in each case.
Problem 3 asks us to compute .
This can be done directly using the chain rule, we get
Problem 4 gives us the graph of for :
The problem then asks to check that indeed has a unique minimum in the interval , and requires that we find it. Then, we must also verify that is concave up.
Note first that the value of requires some explanation. One can check using L’Hôpital’s rule that , so it makes sense to adopt as the value of at 0. To see that the limit is indeed as indicated, note that , so it is enough to check that . Then,
where we can justify the third equality using that the function is continuous.
To compute , we use L’Hôpital’s rule, after rewriting .
Note that and .
Now, and , so
To see that has a unique minimum, we study the for which . We have that
Then precisely when (which cannot happen if , and at we have anyway) or i.e., or .
Recall that is strictly increasing, as can be seen from its graph or from its derivative being :
To determine the concavity of , we examine its second derivative. Note that , so
for all , so is indeed concave up.
Problem 5 asks us to sketch the graph of for . Since it is difficult to solve the equation , the graph of is included below. The solutions to are and .
First we look at extreme points of . For this, we need to consider , and the solutions to .
We have that , so precisely of , i.e., or .
- , and
so gives us a local maximum, gives us a local (in fact, global) minimum, gives us a local (in fact, global) maximum, and gives us a local minimum.
(This can also be checked by analyzing the sign of in the intervals , , , or by looking at the second derivative.)
To determine the concavity of , we consider . The graph is already given to us. We have that if .
In particular, , verifying our conclusion that is a local minimum.
It follows that is concave up in this interval. Also, we see that if .
In particular, , verifying our conclusion that is a local maximum.
This is concave down in this interval, and we also have that is an inflexion point of . Finally, for , so is concave up in this interval, and is another inflexion point.
Combining this with for all , , , , and , we can now draw a rough sketch of the graph of . The program Grapher was used to produce the image below. The lines in blue indicate the points where the concavity of changes.
We could of course have computed directly. We have that , so
To analyze the concavity of , we need to find the zeros of , i.e., we need to solve .
Since , we need
This equation can be solved by combining it with
We have, subtracting from , that
, or .
This equation can be solved using the quadratic formula. We find that
But , and , so . This means that
Unfortunately, there is no closed formula for the corresponding values of with . Using a calculator one can easily verify that the two values of shown above are indeed the solutions to this equation.