How do you find the domain and range of f(x) = (x) / sqrt(x^2+x+1)?

Nov 8, 2017

Domain: $\left(- \infty , \infty\right)$

Range: $\left[- \frac{2}{3} \sqrt{3} , 1\right)$

Explanation:

Note that:

${x}^{2} + x + 1 = {\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4} > 0 \text{ }$ for all real values of $x$

So $\sqrt{{x}^{2} + x + 1}$ is well defined and non-zero for any $x \in \mathbb{R}$

Hence $f \left(x\right)$ is well defined for any $x \in \mathbb{R}$

So its implicit domain is $\mathbb{R}$

To find the range, let:

$y = f \left(x\right) = \frac{x}{\sqrt{{x}^{2} + x + 1}}$

Then:

$y \sqrt{{x}^{2} + x + 1} = x$

Squaring both sides:

${y}^{2} \left({x}^{2} + x + 1\right) = {x}^{2}$

So:

$\left({y}^{2} - 1\right) {x}^{2} + {y}^{2} x + {y}^{2} = 0$

This is a quadratic polynomial equation in $x$ of the form:

$a {x}^{2} + b x + c = 0$

with

$\left\{\begin{matrix}a = {y}^{2} - 1 \\ b = {y}^{2} \\ c = {y}^{2}\end{matrix}\right.$

Its discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c$

$\textcolor{w h i t e}{\Delta} = {\left({y}^{2}\right)}^{2} - 4 \left({y}^{2} - 1\right) {y}^{2}$

$\textcolor{w h i t e}{\Delta} = {y}^{4} - 4 {y}^{4} + 4 {y}^{2}$

$\textcolor{w h i t e}{\Delta} = {y}^{2} \left(4 - 3 {y}^{2}\right)$

When $y = 0$ then $\Delta = 0$ so the quadratic in $x$ has one solution.

Otherwise ${y}^{2} > 0$ so in order for the quadratic in $x$ to have real solutions we require:

$4 - 3 {y}^{2} \ge 0$

That is:

$4 \ge 3 {y}^{2}$

So:

${y}^{2} \le \frac{4}{3}$

and:

$\left\mid y \right\mid \le \sqrt{\frac{4}{3}} = \sqrt{\frac{4 \cdot 3}{9}} = \frac{2}{3} \sqrt{3}$

This is a sufficient condiction for the polynomial to have a solution, but we also require the sign of $x$ to match that of $y$. This is where we run into the limitation of squaring an equation that we did earlier - which can introduce spurious solutions.

Note that in order to have positive solutions, the signs of the coefficients $a$, $b$ and $c$ must differ. Hence we need ${y}^{2} - 1 < 0$. So $\left\mid y \right\mid < 1$.

So the range of $f \left(x\right)$ is $\left[- \frac{2}{3} \sqrt{3} , 1\right)$

graph{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}

Nov 8, 2017

Domain: $\left(- \infty , \infty\right)$

Range: $\left[- \frac{2}{3} \sqrt{3} , 1\right)$

Explanation:

Here's a method using some pre-calculus and calculus...

Note that ${x}^{2} + x + 1 = {\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4} > 0$ for all real values of $x$

Hence for any $x \in \mathbb{R}$ we have:

$\sqrt{{x}^{2} + x + 1} > 0$

Hence $f \left(x\right)$ is well defined for all $x \in \mathbb{R}$

So the domain of $f \left(x\right)$ is $\mathbb{R}$

Note that:

${\lim}_{x \to - \infty} \frac{x}{\sqrt{{x}^{2} + x + 1}} = {\lim}_{x \to - \infty} - \frac{1}{\sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2}} = - 1$

${\lim}_{x \to \infty} \frac{x}{\sqrt{{x}^{2} + x + 1}} = {\lim}_{x \to \infty} \frac{1}{\sqrt{1 + \frac{1}{x} + \frac{1}{x} ^ 2}} = 1$

So there are horizontal asymptotes $y = - 1$ as $x \to - \infty$ and $y = 1$ as $x \to \infty$.

$\frac{d}{\mathrm{dx}} \frac{x}{\sqrt{{x}^{2} + x + 1}} = \frac{d}{\mathrm{dx}} x {\left({x}^{2} + x + 1\right)}^{- \frac{1}{2}}$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \frac{x}{\sqrt{{x}^{2} + x + 1}}} = {\left({x}^{2} + x + 1\right)}^{- \frac{1}{2}} - x \frac{1}{2} {\left({x}^{2} + x + 1\right)}^{- \frac{3}{2}} \left(2 x + 1\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \frac{x}{\sqrt{{x}^{2} + x + 1}}} = {\left({x}^{2} + x + 1\right)}^{- \frac{3}{2}} \left(\left({x}^{2} + x + 1\right) - \frac{1}{2} x \left(2 x + 1\right)\right)$

$\textcolor{w h i t e}{\frac{d}{\mathrm{dx}} \frac{x}{\sqrt{{x}^{2} + x + 1}}} = \frac{x + 2}{2 {\left({x}^{2} + x + 1\right)}^{\frac{3}{2}}}$

So there's one turning point at $x = - 2$

We find:

$f \left(- 2\right) = \frac{- 2}{\sqrt{{\left(- 2\right)}^{2} + \left(- 2\right) + 1}} = - \frac{2}{\sqrt{3}} = - \frac{2}{3} \sqrt{3}$

Hence the range of $f \left(x\right)$ is:

$\left[- \frac{2}{3} \sqrt{3} , 1\right)$

graph{x/sqrt(x^2+x+1) [-5, 5, -2.5, 2.5]}