Gavin Frye

2022-01-29

Range of $f\left(x\right)=\frac{\mathrm{tan}\left\{x\right\}}{\mathrm{tan}\left\{3x\right\}}$
My Attempt:
I wrote down,
$\mathrm{tan}\left\{3x\right\}=\frac{3\mathrm{tan}\left\{x\right\}-{\mathrm{tan}}^{3}\left\{x\right\}}{1-3{\mathrm{tan}}^{2}\left\{x\right\}}$
This reduced f(x) to,
$f\left(x\right)=\frac{1-3{\mathrm{tan}}^{2}\left\{x\right\}}{3-{\mathrm{tan}}^{2}\left\{x\right\}}$
I don't know how to solve any further. I thought of using derivative, but the function is dicontinuous at times.

Jacob Trujillo

Expert

Note that ${\mathrm{tan}\left(x\right)}^{2}\ge 0$ and
$f\left(x\right)=\frac{1-3{\mathrm{tan}\left(x\right)}^{2}}{3-{\mathrm{tan}\left(x\right)}^{2}}=3+\frac{8}{{\mathrm{tan}\left(x\right)}^{2}-3}$
Clearly, for and $f\left(x\right)>3$
For $0\le {\mathrm{tan}\left(x\right)}^{2}<3$, we have
$-3\le {\mathrm{tan}\left(x\right)}^{2}-3<0⇒\frac{8}{{\mathrm{tan}\left(x\right)}^{2}-3}\le -\frac{8}{3}⇒f\left(x\right)=3+\frac{8}{{\mathrm{tan}\left(x\right)}^{2}-3}\le 3-\frac{8}{3}=\frac{1}{3}$

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