Range of $f(x)= \frac{\tan{x}}{\tan{3x}} $

Question

Prove that for the values of $x$ where the following $f(x)$ is defined, $f(x)$ does not lie between $\frac{1}{3}$ and $3$. $$f(x)=\frac{\tan{x}}{\tan{3x}}$$

My Attempt:

I wrote down, $$\tan{3x}=\frac{3\tan{x}-\tan^3{x}}{1-3\tan^2{x}}$$

This reduced $f(x)$ to,

$$f(x)=\frac{1-3\tan^2{x}}{3-\tan^2{x}}$$

I don't know how to solve any further. I thought of using derivative, but the function is dicontinuous at times. How do I solve it? Any hints would be helpful.

Thanks.

Answer 1

Note that $\tan(x)^2 \ge 0$ and $$f(x)= \frac{1-3\tan(x)^2}{3-\tan(x)^2} = 3+\frac{8}{\tan(x)^2-3}.$$ Clearly, for $\tan(x)^2 > 3$, $\frac{8}{\tan(x)^2-3} > 0$, and $f(x) >3$.

For $0 \le \tan(x)^2 < 3$, we have $-3 \le \tan(x)^2-3 < 0 \implies \frac{8}{\tan(x)^2-3}\le-\frac{8}{3} \implies f(x) = 3+\frac{8}{\tan(x)^2-3} \le 3-\frac{8}{3} = \frac{1}{3}.$

Answer 2

Suppose there is an $x$ so that $$1/3\leq \frac{1-3\tan^2{x}}{3-\tan^2{x}}\leq 3$$

so $$\frac{-8}{3-\tan^2{x}}\leq 0\;\;\;\;{\rm and }\;\;\;\;0\leq \frac{-8\tan^2x}{3-\tan^2{x}}$$

so $$0<3-\tan^2x <0$$ which is a contradiction.

Answer 3

\begin{align*} f(x)&=\frac{1-3\tan^2{x}}{3-\tan^2{x}}\\ &=\frac{1-3\frac{\sin^2{x}}{\cos^2{x}}}{3-\frac{\sin^2{x}}{\cos^2{x}}}\\ &=\frac{\cos^2{x}-3\sin^2{x}}{3\cos^2{x}-\sin^2{x}}\\ &=\frac{\cos^2{x}-3(1-\cos^2{x})}{3\cos^2{x}-(1-\cos^2{x})}\\ &=\frac{4\cos^2{x}-3}{4\cos^2{x}-1}\\ &=\frac{4\cos^2{x}-1}{4\cos^2{x}-1}-\frac{2}{4\cos^2{x}-1}\\ &=1-\frac{2}{4\cos^2{x}-1} \end{align*} Now $0\le \cos^2{x}\le1$, so at $\cos^2{x}=0$, $1-\frac{2}{4\cdot0-1}=1-\frac{2}{-1}=3$, and at at $\cos^2{x}=1$ $1-\frac{2}{4\cdot1-1}=1-\frac{2}{3}=\frac{1}{3}$.

To investigate further take the derivative of $f$: $$f'(x)=-\frac{16 \cos x \sin x}{(4 \cos x^2-1)^2}$$ Now at $\cos^2x=0$, we have $\cos x=0$ and so $f'(x)=0$. Similarly at $\cos^2x=1$, we have $\sin x=0$ and so $f'(x)=0$. Hence these are both stationary points for $f(x)$.

To investigate these stationary points take the scond derivative of $f$: $$f''(x)=-\frac{256 \cos^2x \sin^2x}{(4 \cos x^2-1)^3}+ \frac{16 (\sin^2x -\cos^2x)}{(4 \cos x^2-1)^2}$$ Now at $\cos^2x=0$, we have $\sin^2x=1$ and so $f''(x)=16$, implying $f(x)=3$ is a local minimum. Similarly at $\cos^2x=1$, we have $\sin^2x=0$ and so $f''(x)=-16$ implying $f(x)=\frac{1}{3}$ is a local maximum. Putting these two results together shows $f(x)$ cannot have any value between $\frac{1}{3}$ or $3$ as required. (See plot of $f(x)$ below for illustration of result.)

Answer 4

Let $y=\dfrac{1-3t^2}{3-t^2}$

For finite $t, y\ne3$

$\iff3y-yt^2=1-3t^2\iff t^2=\dfrac{3y-1}{y-3}$

For real $t,$ we need $\dfrac{3y-1}{y-3}\ge0$

either $3y-1=0$

or $(3y-1)(y-3)>0$

$\implies$

either $y>$max$\left(\dfrac13, 3\right)=?$

or $y<$min$\left(\dfrac13, 3\right)=?$