Question

$\frac{dy}{dx}= \frac{8cos(x)}{ x^{2} } - \frac{1}{8}$

How many relative maxima and how many relative minima are there on the interval (1,10)? I don't see the way forward so I did some handwaving and said that the function in terms of maxima and minima is driven by cos(x). Since cos(x) has 2 minima and one maximum on that interval, I claimed that the function also must have 2 minima and 1 max.

Is there an analytical way to answer this?

Answer 1

Intermediate value theorem.

Extrema occur at points where $\dfrac{\mathrm dy}{\mathrm dx}=0$, with maxima occurring at $x=c$ if the derivative is positive to the left of $c$ and negative to the right of $c$, and minima in the opposite case.

So suppose you take two values $a,b$. If it turns out that $\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=a}>0$ and $\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=b}<0$, then the IVT guarantees the existence of some $c\in(a,b)$ such that $\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=c}=0$.

Choosing arbitrary values of $a,b$ won't guarantee that exactly one such $c$ exists, though. The function could easily oscillate several more times between $a$ and $b$, intersecting the x-axis more than once, for example. This is where your suspicion can be applied. Knowing that $\cos x=0$ for $x=\dfrac\pi2,\dfrac{3\pi}2,\dfrac{7\pi}2$ (approximately 1.57, 4.71, 7.85, respectively), you can use these values as reference points for computing the sign of the derivative.

When $x=\dfrac\pi2$, you have $\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac18$. You know that $\cos x>0$ for $0<x<\dfrac\pi2$, and that as $x\to0^+$, $\dfrac{\mathrm dy}{\mathrm dx}\to+\infty$. This means there must be some $c\in\left(0,\dfrac\pi2\right)$ such that $\dfrac{\mathrm dy}{\mathrm dx}=0$, and in particular, this value of $c$ is the site of a relative maximum.

You can use similar arguments to determine what happens at the other two suspected critical points.