Name: The Inverse of Cosine Function
Uploaded: 2019-07-19
Duration: 4 min 30 s
Description: The curve of a function and its inverse are mirror images of each other in line y = x, let's draw the graph of the inverse of the cosine function using this...

Question

Find the domain of

f (x) = \sin^{- 1} (\frac{x^{2} + 1}{2 x})

.

Accepted Answer

Given $f (x) = \sin^{- 1} (\frac{x^{2} + 1}{2 x})$

We know that domain of $\sin^{- 1} x is [- 1, 1]$

Now, $- 1 \leq \frac{x^{2} + 1}{2 x} \leq 1$

Consider, $- 1 \leq (\frac{x^{2} + 1}{2 x})$

$\Rightarrow - 1 + 1 \leq \frac{x^{2} + 1}{2 x} + 1$

$\Rightarrow \frac{x^{2} + 1 + 2 x}{2 x} \geq 0$

$\Rightarrow \frac{{(x + 1)}^{2}}{2 x} \geq 0$

${(x + 1)}^{2}$ is non-negative for all values of $x$ .

In order to get $\frac{{(x + 1)}^{2}}{2 x} \geq 0$ we must have $x > 0$

$∴ \frac{{(x + 1)}^{2}}{2 x} \geq 0 \Rightarrow x = - 1 and x > 0$

Consider, $(\frac{x^{2} + 1}{2 x}) \leq 1$

$\Rightarrow \frac{x^{2} + 1}{2 x} - 1 \leq 1 - 1$

$\Rightarrow \frac{x^{2} + 1 - 2 x}{2 x} \leq 0$

$\Rightarrow \frac{{(x - 1)}^{2}}{2 x} \leq 0$

${(x - 1)}^{2}$ is non-negative for all values of $x$ .

In order to get $\frac{{(x - 1)}^{2}}{2 x} \leq 0$ we must have $x < 0$

$∴ \frac{{(x - 1)}^{2}}{2 x} \leq 0 \Rightarrow x = 1 and x < 0$ .

The only values of $x$ satisfying inequalities are $- 1$ and $1$

$∴$ Domain of $f (x)$ is $\{- 1, 1\}$ .

Find the domain of $f (x) = \sin^{- 1} (\frac{x^{2} + 1}{2 x})$ .

Important Questions on Inverse Trigonometric Functions

Learn and Prepare for any exam you want !

Find the domain of fx=sin-1x2+12x.

Important Questions on Inverse Trigonometric Functions

Learn and Prepare for any exam you want !

Find the domain of $f (x) = \sin^{- 1} (\frac{x^{2} + 1}{2 x})$ .