Proving the existence of a root, with an unidentified f(x)

Question

I have the following problem:

1. f(x) is a polynomial function.
2. Prove that the equation: g(x)=tan(x)-f(x) has a root (so g(x) = 0).

I considered the limitations of tan, and tried to get some values to apply the intermediate value theorem but the values do not have opposite signs.

g(-π/4) = -1 - f(-π/4)

g(π/4) = -1 - f(π/4)

g(0) = -f(0)

Am I missing something with what I've found? or my route is wrong?

Answer 1

Restrict your function $g(x)$ in the interval $(-\pi/2, \pi/2)$. Since f is a polynomial hence continuous and hence between $[-M, M]$ for some $M >0$ within the interval $[-\pi/2, \pi/2]$. But $\tan(x) \to \pm \infty$ as $x \to \pm \pi/2$. So you see as you go from $x = -\pi/2$ to $x=\pi/2$ your function $g$ goes from $-\infty$ to $\infty$. By intermediate value theorem (since $g$ is continuous), you have a root of $g$ in $(-\pi/2, \pi/2)$.

It in fact proves $g$ must have an infinite number of roots, at least one in each interval of the form $((2n-1)\pi/2, (2n+1)\pi/2)$ for $n \in \mathbb{Z}$.

Answer 2

Draw a picture. The function $\tan x$ goes from $-\infty$ to $+\infty$ on the interval $(-\pi/2,\pi/2)$. Try to draw the graph of a continuous function on the same interval that doesn't cross $y=\tan x$. By continuity, you'll have to stay above or below the graph of tangent, and your function will have an asymptote at one of $x=\pm\pi/2$. A polynomial can't have this behaviour as it will be continuous on the closed interval $[-\pi/2,\pi/2]$.

Now take the above and try to turn it into something more rigorous. You can use the intermediate value theorem on $p(x)-\tan x$ for values of $x$ near $\pm\pi/2$. E.g. $p(\pi/2-\epsilon)$ and $p(-\pi/2+\epsilon)$ will be close to the values of $p(\pm\pi/2)$ for a polynomial $p$, whereas values of tangent can be made arbitrarily large (positive and negative), so the difference will take on opposite signs.