Finding range of a fraction

Question

Original question was to show that the value of $\frac{\tan x}{\tan 3x}$ wherever defined, newer lies between 1/3 and 3.

So I solved it till here = $ \frac{3t^2-1}{t^2-3}$. Where t = tan x

Now I need to find the range of above function.

I don't have any problems if this was linear or quadratic.

$t^2$ goes from 0 to infinity

$3t^2 - 1$ goes from -1 to infinity

But this is a fraction. How can I solve it??

Also I did try y = f(t) and swapping variables and see where y is defined ( real value) But that gives t = (1/3, 3)

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

Answer 1

Attempt:

$t^2 \not =3:$

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

$(t^2-3)y= 3t^2-1;$

$t^2(y-3) =3y-1;$

$t_{1,2}= \pm \sqrt{\frac{3y-1}{y-3}}$;

1) $3y-1 \ge 0$, and $y>3$;

$y\ge 1/3$, and $y>3$, i.e $y > 3$.

2) $3y-1 \lt 0$, and $y <3$; i.e.$y < 1/3$.

Answer 2

Hint $$\begin{align*} \frac{3t^2-1}{t^2-3}&=\frac{3t^2}{t^2-3}-\frac1{t^2-3}\\&=\frac{3t^2-9}{t^2-3}+\frac8{t^2-3}\\&=\frac{8}{t^2-3}+3\end{align*}$$

What happens when $t\to\pm\infty$ or when $t\to 0$?

Answer 3

For $|t|<3$ we obtain: $$\frac{3t^2-1}{t^2-3}=\frac{1}{3}-\frac{8t^2}{3(3-t^2)}\leq\frac{1}{3}.$$ Fot $|t|>3$ see the Dr. Mathva's post.