Question

find the domain of f(x)=sec^-1 2x​

Answer 1

We have to find the domain of,

$f(x)=\sec^{-1}(2x)$

We know $\cos\theta\in[-1,\ 1].$

This gives  $\sec\theta\in(-\infty,\ -1]\cup[1,\ \infty).$

Therefore,  $2x\in(-\infty,\ -1]\cup[1,\ \infty)$

$\implies \boxed{x\in\left(-\infty,\ -\dfrac{1}{2}\right]\cup\left[\dfrac{1}{2},\ \infty\right)}$

Hence the domain is got.

Answer 2

Answer:

$\mathrm{Domain\:of\:}\:\arcsec \left(2x\right)\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-\dfrac{1}{2}\quad \mathrm{or}\quad \:x\ge \dfrac{1}{2}\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-\dfrac{1}{2}\right]\cup \:\left[\dfrac{1}{2},\:\infty \:\right)\end{bmatrix}$

Step-by-step explanation:

$\blue{\mathrm{Domain\:de finition}}$

• The domain of a function is the set of input or argument values for which the function is real and defined

$\green{\mathrm{Find\:known\:functions\:domain\:restrictions}:}$

$\gray{\arcsec \left(f\left(x\right)\right)\quad \Rightarrow \quad \:f\left(x\right)\le \:-1\quad \mathrm{or}\quad \:f\left(x\right)\ge \:1}$

$\black{\mathrm{Solve\:}\:2x\le \:-1\quad \mathrm{or}\quad \:2x\ge \:1:}$

$\gray{\mathrm{Solve\:}\:2x\le \:-1:\quad x\le \:-\dfrac{1}{2}}$

$\gray{\mathrm{Solve\:}\:2x\ge \:1:\quad x\ge \dfrac{1}{2}}$

$x\le \:-\dfrac{1}{2}\quad \mathrm{or}\quad \:x\ge \dfrac{1}{2}$

$\gray{\mathrm{The\:function\:domain}}$

$x\le \:-\dfrac{1}{2}\quad \mathrm{or}\quad \:x\ge \dfrac{1}{2}$