Question

Find the value(s) of x for the relation 2y = (x+1)(x-1) at f(x) = 1 and determine whether it a function or not. Given that $x$ is an independent variable and $y$ is a dependent variable. Answer in Yes/No.

Answer 1

The given relation is $2y=(x+1)(x-1).$
Given: $f\left(x\right)=1$
$\therefore$ We need to evaluate $x$ at $y=1.$
$2\left(1\right)=(x+1)(x-1)$
$\Rightarrow 2=x\cdot x+x-x-1$
$\Rightarrow 2+1=x^2$
$\Rightarrow x^2=3$

Here, $x$ can have two values.
For $x=\sqrt{3},-\sqrt{3}$

Here, $x$ is independent variable, having two sets of values that results to a single value, i.e., $f\left(x\right)=1.$

$\therefore$ This relation is a function.