Range of $x \mapsto \frac{x+5}{\sqrt{x^2+1}}$

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.

• As a tip, you can always check your answer for example with WolframAlpha or plot the function. That is of course not a mathematical proof, but it helps to get a better understanding. Commented Aug 11, 2021 at 6:46

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 $$f(x) \in (-1, \sqrt{26}]$$.

• You got the derivative incorrectly. Also when you computed the limit at $+\infty$, the function you consider is not the right one.
– Gary
Commented Aug 11, 2021 at 6:42
• @Gary Thank you! Typo, sorry, lost track of the 5 Commented Aug 11, 2021 at 6:45
• The derivative is now correct, but you still have a typo in the limit. The result is nevertheless correct Commented Aug 11, 2021 at 6:47
• OH MY! Hopefully everything is fixed now :) Commented Aug 11, 2021 at 6:48
• To be fair, I did just finish a solid day of research -- I'm quite fried mentally at the moment (possible too fried for MSE) :) Commented Aug 11, 2021 at 6:49