Find the range of the following function $$\begin{aligned} f : \Bbb R &\to \Bbb R\\ x &\mapsto \frac{x+5}{\sqrt{x^2+1}}\end{aligned}$$
I tried squaring both sides and got a quadratic equation in $x$. Then I applied $D\geq0$ as the domain of given function is $x\in \mathbb R$ but by this method answer was $-\sqrt{26} \leq f(x) \leq \sqrt{26}$ but this is not correct answer can anyone please tell where am I doing wrong.
Best Answer
The function $x^2 + 1$ is never zero, hence, $\frac{1}{\sqrt{x^2+1}}$ is defined for all $x \in \mathbb{R}$. The function $x + 5$ is obviously defined for all $x \in \mathbb{R}$, hence the ratio is well-defined for all $x \in \mathbb{R}$. Further, it is a ratio of continuous functions, and therefore continuous.
It suffices to check the boundary limits and interior stationary points.
Boundary limits yield $$\lim_{x \to \infty} \frac{x+5}{\sqrt{x^2+1}} = 1, \hspace{1cm} \lim_{x \to -\infty} \frac{x+5}{\sqrt{x^2+1}} = -1.$$
The stationary points are given by $$f'(x) =0 \implies \frac{1-5x}{(x^2+1)^{\frac{3}{2}}} = 0 \implies x = \frac{1}{5},$$ and $f(1/5) = \sqrt{26}$.
The range is therefore $f(x) \in (-1, \sqrt{26}]$.