At what value(s) of $x$ does $ \cos(x) = 7x$? For some reason, this question was posed to me on a calculus quiz, and I have completely drawn a blank on how to solve this problem.
If someone could explain how to solve this problem, it would be greatly appreciated.
Thanks.
At what value(s) of $x$ does $\cos(x) = 7x$?

We can approximate a solution $a$ of the equation $\cos x = 7x$ using Newton's Method. First observe that a solution to this is the same as a zero of the function $f(x) = 7x - \cos x$. So we'll apply Newton's method to $f(x)$, but we'll need to begin with a good starting value $x_0$.
Just by thinking about how the functions $y = 7x$ and $y = \cos x$ grow, you should be able to see immediately that any and all zeros of $f(x)$ must occur near $x = 0$. Indeed, since $-1 \leq -\cos x \leq 1$ for all $x$, you know that $7x - 1 \leq f(x) = 7x - \cos x \leq 7x + 1$. So if, for example, $x > 1/7$ you know that $0 = 7(1/7) - 1 < 7x - 1 < f(x)$, and similarly $f(x) < 0$ if $x < - 1/7$. So a safe bet for $x_0$ would be $x_0 = 0$ or $x_0 = 1$. Applying the formula $x_{n+1} = x_n - f(x_n)/f'(x_n)$ repeatedly will give you better and better approximations of one solution to this equation. (Now a question for you is: Is there only one such solution? Why?)
