MATH100 December 2013
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Question 11

FullSolution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise.
Prove that there exists at most one positive real number $\displaystyle x$ such that $\displaystyle 4^{x}=3\cos x$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

Consider the function
 $\displaystyle f(x)=4^{x}3\cos x$
How can you relate the problem to the properties of this function?

Hint 2

Show that $\displaystyle f(x)$ has at most one positive zero.

Hint 3

Use monotonicity to show that the positive zero of $\displaystyle f(x)$ must be unique (if it exists).

Hint 4

Note that it is not necessary to show that a solution exists. (Although this can be done using the Intermediate Value Theorem.)

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution

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First we observe that $\displaystyle 4^{x}$ and $\displaystyle 3\cos x$ are both continuous on $\mathbb {R}$. It follows that $\displaystyle f(x)=4^{x}3\cos x$ is also continuous on $\mathbb {R}$. Note that $f(x)=0\iff 4^{x}=3\cos x$. We must show that $\displaystyle f(x)$ has at most one zero.
Optional: Show that a zero exists.
Plugging in the values $\displaystyle f(0)=4^{0}3\cos(0)=13=2$ and $f\left({\frac {\pi }{2}}\right)=4^{\pi /2}3\cos \left({\frac {\pi }{2}}\right)=4^{\pi /2}$ we see that $\displaystyle f(0)<0<f\left({\frac {\pi }{2}}\right)$. Thus by the Intermediate Value Theorem, $\exists c\in \left(0,{\frac {\pi }{2}}\right):f(c)=0.$ In particular, this implies that this positive number c satisfies $\displaystyle 4^{c}=3\cos c.$
Necessary: Show that the zero is unique.
If $\displaystyle f(x)$ is monotonically increasing (or monotonically decreasing) for all positive values, then $\displaystyle f(x)$ can only have one zero in the interval $\displaystyle (0,\infty )$. Hence we calculate
 $\displaystyle f'(x)=4^{x}\ln 4+3\sin x.$
$\displaystyle f(x)$ is increasing where $\displaystyle f'(x)>0$, i.e., where $\displaystyle 4^{x}\ln 4>3\sin x$. We consider the positive real numbers as follows:
Case I: $\displaystyle 0<x<\pi$. Since $\displaystyle 4^{x}\ln 4>0$ and $\displaystyle 3\sin x<0$ on this interval, $\displaystyle 4^{x}\ln 4>3\sin x$ on the interval $\displaystyle (0,\pi )$.
Case II: $x\geq \pi$. Since $3\leq \displaystyle 3\sin x\leq 3$ and $\displaystyle 4^{x}\ln 4\geq 4^{\pi }\ln 4>3\ln e=3$ on this interval, $\displaystyle 4^{x}\ln 4>3\sin x$ on the interval $\displaystyle [\pi ,\infty )$.
It follows that $\displaystyle f'(x)>0$ on the interval $\displaystyle (0,\infty )$ and that $\displaystyle f(x)$ is monotonically increasing on the interval $\displaystyle (0,\infty )$. We therefore conclude that if there exists a positive number c such that $\displaystyle 4^{c}=3\cos c$, this number must be unique.

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