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  4. The range of the function f(x)=(x2+8/x2+4), x ∈ R is

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Q. The range of the function $f\left(x\right)=\frac{x^{2}+8}{x^{2}+4}$, $x \in R$ is

Relations and Functions

Solution:

$f\left(x\right)=\frac{x^{2}+8}{x^{2}+4}$
$\Rightarrow f\left(x\right)=\frac{x^{2}+4}{x^{2}+4}+\frac{4}{x^{2}+4}$
$\Rightarrow f\left(x\right)=1+\frac{4}{x^{2}+4} > 1$
$\Rightarrow f\left(x\right)$ is maximum at $x=0$.
Hence $R\left(f\left(x\right)\right)=(1, 2]$