How to find the domain and the range of this function$f(x)=\sqrt{5-\frac{x^2}{x^2+2}}$ algebraically?

Question

help how to find the domain and the range of this function algebraically

$$f(x)=\sqrt{5-\frac{x^2}{x^2+2}}$$

Answer 1

It amounts to finding the domain and range of $f(x) = \dfrac{x^2}{x^2+2}$. Clearly the domain is all $\mathbb{R}$, and for the range, $0 \le f(x) < 1\implies 0 \ge -f(x) > -1\implies \sqrt{5-\dfrac{x^2}{x^2+2}}= \sqrt{5-f(x)}\in(2,\sqrt{5}].$

Answer 2

Hint:

There are two separate conditions to consider for $f(x)$ to be defined.

$$f(x)=\sqrt{5-\frac{x^2}{x^2+2}}$$

• The radicand must be non-negative. Hence,

$$5-\frac{x^2}{x^2+2} \geq 0$$

• The denominator shouldn't be equal to $0$.

$${x^2+2} \neq 0$$

Since $x^2 \geq 0$ for all real values of $x$, the second condition is already true.

You can find the range after finding the domain.

Answer 3

hint

$$5-\frac{x^2}{x^2+2}=2(2+\frac{1}{x^2+2})\ge 0$$ the domain is $\Bbb R$.

Answer 4

Let's make the bijective change of variable $\quad x=\sqrt{2}\sinh(u)\quad$ with $u\in\mathbb R$.

Then $\displaystyle f(x)=\sqrt{5-\frac{x^2}{x^2+2}}=\sqrt{4+\frac 1{\cosh(u)^2}}$

Since $\cosh(u)\in [1,+\infty[$, we can see that $f$ is defined everywhere (denominator is not zero and what's inside the square root is positive).

The range of $\dfrac 1{\cosh(u)^2}$ being $]0,1]$ then we can deduce the range of $f$ to be $]2,\sqrt{5}]$.

Answer 5

$5-\dfrac{x^2+2-2}{x^2+2}=$

$4+\dfrac{2}{x^2+2}.$

$4 \lt 4+\dfrac{2}{x^2+2} \le 5$, $x \in \mathbb{R}.$

Domain: $\mathbb{R}$.

Range: $(2,√5]$.