# Enrichment, Extension, and Application

1 point

## Exercises

1. Derivative Play
1. Find the derivative of each function, then show that they are equal.
1. $$f(x) = \sin(2x)\cos(2x)$$
2. $$g(x) = \frac{1}{2}\sin(4x)$$
2. For $$f(x) = \dfrac{\sin^2(x)}{1 + \cot(x)} + \dfrac{\cos^2(x)}{1 + \tan(x)}$$, show that $$f'(x) = -\cos(2x)$$.
3. Use implicit differentiation to find $$y'(1)$$ if $$y(x)$$ is defined by the equation $x^3 + \tan(y) = 2x$ for $$-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$$.
4. Find all the critical points of the function $$f(x) = \cos\left( x + \frac{1}{x} \right)$$.
5. Use logarithmic differentiation to find each of the following derivatives:
1. $$y'(x)$$ for $$y = x^{\ln(x)}$$ on $$x \gt 0$$
2. $$y'\left( \frac{\pi}{2} \right)$$ for $$y = x^{\sin(x)}$$
2. Limit Play
1. Note: If you have not studied the fundamental trigonometric limit $$\displaystyle \lim_{\theta \to 0} ~ \dfrac{\sin(\theta)}{\theta}$$ before, please take some time to carefully read the Appendix included at the end of this problem set before proceeding.

Recall that the limit $$\displaystyle ~ \lim_{h \to 0} ~ \dfrac{\cos{(h)} - 1}{h}$$ is needed to prove that $$\dfrac{d}{dx} \Big( \sin(x) \Big) = \cos(x)$$. Use the substitution $$h = 2u$$ and the trigonometric identity $$\cos{(2u)} = 1 - 2\sin^2{(u)}$$ to rewrite this limit, and hence prove that it equals $$0$$.

2. Using algebraic manipulation and the fact that $$\displaystyle \lim_{x \to 0} ~ \dfrac{\sin(x)}{x} = 1$$, evaluate each of the following limits:
1. $$\displaystyle \lim_{x \to 0} ~ \dfrac{x^2 (3 + \sin(x))}{(x + \sin(x))^2}$$
2. $$\displaystyle \lim_{x \to 0} ~ \dfrac{\tan(x) \sin(2x)}{x^2}$$
3. $$\displaystyle \lim_{u \to \infty} ~ u \sin{\left( \dfrac{1}{u} \right)}$$ (Hint: Let $$x = \dfrac{1}{u}$$; what happens to $$x$$ as $$u \to \infty$$?)
3. From Graphs to Roots
1. Using suitable function graphs and their properties, show that the function $f(x) = x^2 - \cos(x)$ satisfies $$f(x) = 0$$ for exactly two real numbers $$a$$ and $$b$$. State an interval containing $$a$$, and one containing $$b$$.
2. Find a root of $$f(x) = \left\lvert \sin(2x) \right\rvert - \ln(x) - 1$$ as follows:
1. Sketch the two graphs $$y = \left\lvert \sin(2x) \right\rvert$$ and $$y = \ln(x) + 1$$ on the same axes over the interval $$-\pi \lt x \lt \pi$$, showing all intercepts.
2. Explain, using the sketch from i), how you know that there is a number $$b$$ in $$(0, 1]$$ such that $$f(b) = 0$$.
3. Use Newton's method to find the value of $$b$$ accurate to $$4$$ decimal places.
4. Getting in the Swing of Things

A runner's elbow swings back and forth from the shoulder as he runs. This motion is illustrated in the diagram shown, where point $$O$$ is located at the runner's shoulder and point $$P$$ is located at the runner's elbow. If $$y$$ is the angle between a vertical line passing through $$O$$ and the line segment $$OP$$, then the motion of the elbow at $$P$$ can be described by $y(t) = \frac{\pi}{8} \cos{(3\pi t)}$ where $$t$$ is in seconds.

1. Sketch a graph of $$y(t)$$ for $$0 \leq t \leq 2$$, and state the amplitude $$A$$ and the period $$T$$ of the motion. Label each point on your graph where $$y(t) = A$$.
2. If the distance $$OP$$ from the shoulder to the elbow is $$30$$ cm, what is the maximum value of the horizontal displacement $$h$$ of the elbow?
3. Assuming the runner takes $$2$$ steps per period of the elbow's swing, how many steps per minute does he take?
4. When someone decides to walk instead of run, how would the amplitude $$A$$ and period $$T$$ of $$y(t)$$ change?
5. Can you ‘induct’ this?

The following problem requires an understanding of the method of mathematical induction. Ask a math educator or perform a web search for “mathematical induction” if you are unfamiliar with the concept.

Given $$f(x) = \ln(x - 1)$$ for $$x \gt 1$$, show that $f^{(n)}(x) = \dfrac{(-1)^{n - 1} (n - 1)!}{(x - 1)^n}$

6. The Invertibles!
1. Consider the positive function $$f(x) = \dfrac{1}{1 + e^x}$$.
1. Determine the horizontal asymptotes of $$y = f(x)$$ by evaluating $$\displaystyle \lim_{x \to \infty} f(x)$$ and $$\displaystyle \lim_{x \to -\infty} f(x)$$.
2. Is $$f(x)$$ an even function, an odd function, or neither?
3. Find $$f'(x)$$ and $$f''(x)$$, and hence determine any relative extremes and/or points of inflection of $$y = f(x)$$.
4. Explain why $$f$$ has an inverse $$f^{-1}$$. Determine the inverse function $$f^{-1}(x)$$, and state its domain and range.
5. Sketch the graphs $$y = f(x)$$, $$y = x$$, and $$y = f^{-1}(x)$$ on the same set of axes, clearly indicating all asymptotes, relative extremes, and/or points of inflection of both $$y = f(x)$$ and $$y = f^{-1}(x)$$.
2. Repeat parts i) to v) of part a) for the function $$f(x) = \ln{\left( x + \sqrt{x^2 + 1} \right)}$$. (Hint: Show that $$\left( x + \sqrt{x^2 + 1} \right)\left(-x + \sqrt{x^2 + 1} \right) = 1$$.)
7. Squeezing Out Results!
1. Suppose that a function $$f(x)$$ is defined as follows: $f(x) = \begin{cases} x^2 \sin{\left( \frac{\pi}{x} \right)} & x \neq 0 \\ 0 & x = 0 \end{cases}$
1. Show that $$f(x)$$ is continuous at $$x = 0$$ by proving that $$f(0) = \displaystyle \lim_{x \to 0} ~ f(x)$$.

(Hint: Try using the squeeze theorem with $$m(x) = -x^2$$ and $$M(x) = x^2$$.)

2. Show that $$f'(0) = \displaystyle \lim_{x \to 0} ~ \dfrac{f(x) - f(0)}{x}$$ exists.

(Hint: What squeeze theorem functions $$m(x)$$ and $$M(x)$$ do you need here?)

3. Is $$f(x)$$ continuously differentiable at $$x = 0$$, i.e. is $$f'(x)$$ continuous at $$x = 0$$?

(Hint: Recall that a function $$F$$ is continuous at $$x = a$$ if $$\displaystyle \lim_{x \to a} F(x) = F(a)$$.)

4. It is a commonly held belief that if $$f'(c) = 0$$, then $$x = c$$ must be a local extreme or a point of inflection of $$f(x)$$. Think about whether $$f(x)$$ satisfies this behaviour at $$x = 0$$.

(Hint: Examine the behaviour of $$f(x)$$ near $$x = 0$$. What happens at the points $$x = \dfrac{2}{4k + 1}$$ versus $$x = \dfrac{2}{4k + 3}$$, where $$k$$ is any integer?)

2. Suppose that a function $$g(x)$$ is defined as follows: $g(x) = \begin{cases} x^4 \sin^2{\left( \frac{\pi}{x} \right)} & x \neq 0 \\ 0 & x = 0 \end{cases}$
1. Explain how you know that $$g(0)$$ must be a local minimum of $$g(x)$$.
2. Show that both $$g'(0) = 0$$ and $$g''(0) = 0$$.

(Hint: The results proven in part a) will be useful.)

3. Below are the graphs $$y = f(x)$$ and $$y = g(x)$$. Which graph is which? Justify your answer in as many ways as you can.
8. Maximums by the Back Door
1. Find the maximum value of the function $$g(x) = \dfrac{\ln(x)}{x}$$ on $$x \gt 0$$.
2. Without further calculation, find the maximum value of $$f(x) = x^{\frac{1}{x}}$$ on $$x \gt 0$$. Justify your reasoning carefully.
1. By finding the minimum of a single-variable function, show that $\dfrac{e^{x + y}}{xy} \geq e^2$ for all $$x, y \gt 0$$.
9. Potpourri
1. A function $$f$$ satisfies $$f(\pi) = 0$$, and $$f'(x) = \cos{\Big( \cos(x) \Big)}$$ for all real $$x$$. Is the value of $$f(0)$$ positive, negative, or $$0$$?

(Hint: What does the mean value theorem say about $$f(\pi) - f(0)$$?)

2. Show that $$x - \frac{1}{2}x^2 \lt \ln{(1 + x)} \lt x$$ for all $$x \gt 0$$.

(Hint: First graph the three functions on the same axes, then compare their derivatives on $$x \gt 0$$.)

10. Extension: An Inverse Trigonometric Function

We know how to find the tangent of a given angle; however, sometimes it is useful to have a function that outputs the angle, given the tangent ratio — that is, the inverse of the tangent operation.

Specifically, given $$f(x) = \tan(x)$$, we would like to find a function $$f^{-1}(x)$$ such that $$f^{-1}(x) =$$ the angle $$\theta$$ whose tangent is $$x$$, so that $f^{-1} \Big(f(x)\Big) = f^{-1} \Big( \tan(x) \Big) = x$ for suitable real values of $$x$$.

Recall that an inverse for some function $$f(x)$$ is generally found by first interchanging $$x$$ and $$y$$ in the equation $$y = f(x)$$ and then solving algebraically for $$y$$.

For example, if $$y = f(x) = e^x$$ for any real $$x$$, then interchanging $$x$$ and $$y$$ gives $$x = e^y$$, which can be solved to give $$y = \ln(x)$$, but only for $$x \gt 0$$. Thus, the function $$f(x) = e^x$$ has inverse $$f^{-1}(x) = \ln(x)$$. Note that $$f^{-1}\Big( f(x) \Big) = \ln{\left( e^x \right)} = x$$ for all real $$x$$, and $$f\Big( f^{-1}(x) \Big) = e^{\ln(x)} = x$$ for all $$x \gt 0$$; each function reverses the effect of the other on its own domain.

The algebraic operation of interchanging $$x$$ and $$y$$ corresponds to the geometric reflection of the curve $$y = f(x)$$ in the line $$y = x$$, giving the curve $$x = f(y)$$, or $$y = f^{-1}(x)$$. Applying this transformation to the graph of $$y = \tan(x)$$ in the line $$y = x$$ gives Figure A:

However, Figure A has a problem: the reflected graph is not the graph of a function, which must have a single $$y$$-value for every $$x$$-value — this is not the case in the diagram. It is actually impossible to define an inverse function for a periodic function across its entire domain. Instead, we must define an inverse on a subset of the domain of such a function. Specifically, we must choose a subset on which the function is one-to-one.

For the tangent function, the convention is to define the inverse on an interval centred at $$x = 0$$, i.e., the interval $$-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$$.

Thus, for the tangent function, while $$x = \tan(y)$$ depicts the multiple-branched graph portrayed in Figure A, we focus on the middle branch only (sometimes called the principal branch) and label that the inverse tangent function $$y = \tan^{-1}(x)$$, or more often, $$y = \arctan(x)$$. The notation “$$\arctan(x)$$” is favoured over “$$\tan^{-1}(x)$$” in order to avoid confusion with the reciprocal function $$(\tan(x))^{-1} = \cot(x)$$.

The notation $$y = \arctan(x)$$ can be read as “$$y$$ is the angle whose tangent is $$x$$.” A graph of $$\tan(x)$$ on $$-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$$ and its inverse $$y = \arctan(x)$$ are shown:

Notice that, as was the case with inverse functions $$e^x$$ and $$\ln(x)$$, the domains and ranges are reversed. In the case of $$\tan(x)$$, the chosen domain of $$\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$$ of $$\tan(x)$$ becomes the range of $$\arctan(x)$$, while the range $$(-\infty, \infty)$$ of $$\tan(x)$$ becomes the domain of $$\arctan(x)$$. Furthermore, the vertical asymptotes $$x = \pm \frac{\pi}{2}$$ of $$\tan(x)$$ become the horizontal asymptotes $$y = \frac{\pi}{2}$$ as $$x \to \infty$$, and $$y = -\frac{\pi}{2}$$ as $$x \to -\infty$$ of $$\arctan(x)$$.

The following are some problems to enrich understanding of inverse trigonometric functions:

1. Evaluate each expression.
1. $$\sin{\big( \arctan{(\sqrt{3})} \big)}$$
2. $$\arctan{\big( \sin{\left( \frac{\pi}{6} \right)} \big)}$$
3. $$\arctan{\big( \tan{\left( \frac{5\pi}{6} \right)} \big)}$$ (Why is it not $$\frac{5\pi}{6}$$?)
4. $$\cos{\big( \arctan{(3)} \big)}$$
5. Show that $$\cos{\big( \arctan(x) \big)} = \dfrac{1}{\sqrt{1 + x^2}}$$ for all real $$x$$.

(Hint: for parts i), iv), and v), sketch a suitable right angled triangle and label the relevant angle $$\theta$$; for example, in i), the triangle will have sides $$1$$ and $$\sqrt{3}$$, and $$\theta$$ will be $$\arctan{(\sqrt{3})}$$.)

2. The derivative of $$\arctan(x)$$ is $\dfrac{d}{dx} \Big( \arctan(x) \Big) = \dfrac{1}{1 + x^2}$ (This result is proven in problem 11 below.) Use this formula, and the chain rule, to find the derivative of each of the following functions:
1. $$\sin{\big(\arctan(x)\big)}$$
2. $$\arctan{\left( \dfrac{x}{1 + x} \right)}$$
3. $$e^{g(x)}$$ where $$g(x) = \arctan{\left( x^2 \right)}$$
3. Show that $$f(x) = \arctan{\left( \dfrac{ax - 1}{ax + 1} \right)} - \arctan(ax)$$, for $$a \gt 0$$, is a constant function on the interval $$x \gt -\frac{1}{a}$$. Then, find the constant value that $$f(x)$$ obtains on this interval.
4. Use the mean value theorem to show that $$\frac{1}{10} \lt \arctan{(3)} - \arctan{(2)} \lt \frac{1}{5}$$.
11. Bridges and Boats

The Puente de la Mujer in Buenos Aires, Argentina, designed by Santiago Calatrava, is a unique bridge divided into three sections. The two shorter side sections, joined to land, are fixed in place, while the centre section is actually a swing bridge, which can be rotated 90 degrees to allow ships to pass by. The swing bridge portion is held up by cables attached to the white pylon in the centre of the bridge, as shown in the diagram:

One interesting question to ask is: if you approached the bridge in a boat, hoping to get the best possible photograph, from how far away should the photo be taken?

The tip of the cantilever lies $$34$$ metres above the bridge. The best view of an object can be characterized as when the angle $$\theta$$ subtended at the eye is maximized. Assuming the bridge itself is about $$6$$ metres above the eye level of an observer in the boat, then the situation can be modelled by the diagram shown to the right.

From the diagram, the quantity $$\theta$$ can be described as the difference of two angles, as follows: $\theta = (\theta + \alpha) - \alpha = \left( \text{the angle whose tangent is}~\frac{40}{x} \right) - \left( \text{the angle whose tangent is}~\frac{6}{x} \right)$

Using the $$\arctan$$ function, the viewing angle $$\theta$$ can be written as a function of the distance $$x$$ to the bridge as follows: $\theta = (\theta + \alpha) - \alpha = \arctan{\left( \frac{40}{x} \right)} - \arctan{\left( \frac{6}{x} \right)}$

However, in order to maximize the viewing angle $$\theta$$ with respect to $$x$$, we need to be able to find the derivative of the $$\arctan$$ function. This can be found using implicit differentiation.

We know that $$y = \arctan(x)$$ means that $$x = \tan(y)$$ for $$-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}$$. Differentiating $$x = \tan(y)$$ implicitly with respect to $$x$$ we get $1 = \sec^2(y) \dfrac{dy}{dx}$ giving $\dfrac{dy}{dx} = \dfrac{1}{\sec^2(y)} = \dfrac{1}{1 + \tan^2(y)} = \dfrac{1}{1 + x^2}$ and thus the required derivative is $\dfrac{d}{dx} \Big( \arctan(x) \Big) = \dfrac{1}{1 + x^2}$

Applying the chain rule to $$\theta = \arctan{\left( \dfrac{40}{x} \right)} - \arctan{\left( \dfrac{6}{x} \right)}$$, \begin{align*} \dfrac{d\theta}{dx} &= \left( \dfrac{1}{1 + \left( \frac{40}{x} \right)^2} \right) \left( - \dfrac{40}{x^2} \right) - \left( \dfrac{1}{1 + \left( \frac{6}{x} \right)^2} \right) \left( - \dfrac{6}{x^2} \right) \\ &= \dfrac{-40}{x^2 + 40^2} - \dfrac{-6}{x^2 + 6^2} \\ &= \dfrac{8160 - 34x^2}{\left( x^2 + 40^2 \right)\left( x^2 + 6^2 \right)} \end{align*}

Setting $$\dfrac{d\theta}{dx}$$ equal to $$0$$ and solving, the only positive critical value is $$x = \sqrt{\frac{8160}{34}} = \sqrt{240} \approx 15.5$$ metres. In fact, since $$\dfrac{d\theta}{dx}$$ is positive when $$x \lt \sqrt{240}$$ and negative when $$x \gt \sqrt{240}$$, the function must have a maximum value at that critical value.

Hence, the optimum distance to take a photo is at a distance of about $$15.5$$ metres away from the bridge.

A similar problem is as follows:

Two smugglers in separate boats are floating along a river. A police camera is hidden $$100$$ m away from a point $$P$$ directly downstream from the smugglers' position (in a direction perpendicular to the river's flow). Both boats pass $$P$$ at the same time, one travelling three times as fast as the other. As they pass, the camera rotates and follows their path, taking a photo when the viewing angle subtending the distance between the two boats is at a maximum. (Something to think about: is this really the optimal viewing angle for a police photo?)

How far from the slower boat is the faster boat when the photo is taken?

### Appendix: The Fundamental Trigonometric Limit

You may have seen numerically that $$\displaystyle \lim_{\theta \to 0} ~ \dfrac{\sin(\theta)}{\theta}$$ appears to have a value of $$1$$. A more formal proof can be done using geometry and the squeeze theorem (which has been introduced in previous enriched problem sets).

Assume initially that $$0 \lt \theta \lt \frac{\pi}{2}$$. In the diagram shown, the point $$C$$ on the unit circle has coordinates $$\big( \cos(\theta), \sin(\theta) \big)$$. The length of $$CD$$ equals the $$y$$-coordinate of $$C$$, which is $$\sin(\theta)$$.

By trigonometric ratios, $$\dfrac{BA}{OA} = \dfrac{BA}{1} = \tan(\theta)$$, so $$BA = \tan(\theta)$$. Finally, the length of arc $$AC$$ is $$r\theta = \theta$$, since we are dealing with a unit circle.

From the diagram, we note the relationship between the lengths of line segments $$CD$$ and $$BA$$ versus the length of the arc $$AC$$: since the line segment between $$A$$ and $$C$$ cannot be of greater length than arc $$AC$$ (a line being the shortest distance between two points), and the length of $$CD$$ is less than that of the line from $$A$$ to $$C$$ (the latter being the hypotenuse of the triangle $$\triangle{DAC}$$), it follows that $$CD$$ must be shorter than arc $$AC$$.

It can also be shown that the length of arc $$AC$$ is less than that of line segment $$BA$$, for $$0 \lt \theta \lt \frac{\pi}{2}$$, though such a proof is beyond the scope of this course.

Thus, using these facts and noting that $$CD = \sin(\theta)$$, arc $$AC = \theta$$, and $$BA = \tan(\theta)$$, we can write $\sin(\theta) \lt \theta \lt \tan(\theta)$ Since $$0 \lt \theta \lt \frac{\pi}{2}$$, then $$\sin(\theta) \gt 0$$ and $$\cos(\theta) \gt 0$$. So we may divide the inequality by $$\sin(\theta)$$ to obtain $1 \lt \dfrac{\theta}{\sin(\theta)} \lt \dfrac{1}{\cos(\theta)}$ Taking the reciprocal of each term (and hence reversing the inequalities), we get $\cos(\theta) \lt \dfrac{\sin(\theta)}{\theta} \lt 1$

Consider what happens in the limit as $$\theta \to 0^{+}$$. Since $$\cos(\theta) \to 1$$ in this case, and the right-hand term remains $$1$$, the middle term $$\dfrac{\sin(\theta)}{\theta}$$ is "squeezed" towards a value of $$1$$ as $$\theta \to 0^{+}$$.

This reasoning is justified by recalling the squeeze theorem, which states that if there exist functions $$m(x)$$, $$f(x)$$, and $$M(x)$$, which

• are all defined near $$x = a$$ (that is, at every real number within some interval around $$a$$, except possibly at $$x = a$$),
• satisfy $$m(x) \leq f(x) \leq M(x)$$ at these points, and
• $$\displaystyle \lim_{x \to a} m(x) = L = \lim_{x \to a} M(x)$$,

then $$\displaystyle \lim_{x \to a} ~ f(x) = L$$ as well.

The diagram illustrates visually how such a "squeezing" process works with $$x = \theta$$, $$a = 0$$, $$m\left( \theta \right) = \cos(\theta)$$, $$M\left( \theta \right) = 1$$, and $$f\left( \theta \right) = \dfrac{\sin(\theta)}{\theta}$$.

Note that $$f\left( -\theta \right) = f\left( \theta \right)$$ for all $$\theta$$ near $$0$$, so $$\displaystyle \lim_{\theta \to 0^{-}} f(\theta) = \lim_{\theta \to 0^{+}} f(\theta)$$ and the limit exists. Therefore, since $$\cos(\theta) \lt \dfrac{\sin(\theta)}{\theta} \lt 1$$ for all values of $$\theta$$ near $$0$$, the squeeze theorem implies that $\lim_{\theta \to 0^{+}} \dfrac{\sin(\theta)}{\theta} = 1$