If \[y=\frac{ax}{x^2+x+1} \qquad (a>0)\] show that for real values of \(x\) \[-a\leq y \leq \frac{1}{3}a.\]
Multiplying through by the denominator gives \[x^2y+(y-a)x+y = 0,\] which we can think of as a quadratic in \(x\).
To ensure that \(x\) is always real, we need the discriminant to be non-negative, which means \[\begin{align*} 0 &\le (y-a)^2-4y^2 \\ \iff 0 &\le y^2 -2ay+a^2-4y^2 \\ \iff 0 &\le -3y^2 -2ay+a^2 \\ \iff 0 &\ge 3y^2 +2ay-a^2 \\ \iff 0 &\ge (3y-a)(y+a). \end{align*}\]
From the sketch, we can see that this inequality is satisfied when \[-a \le y \le \frac{a}{3}\] as required.
Sketch the graph of \(y\) in the case \(a=6\).
Firstly, we can check for vertical asymptotes occurring when the denominator is \(0\). \[x^2+x+1=0 \Longrightarrow x = \frac{-1\pm \sqrt{-3}}{2}.\] These roots are not real, so the denominator of \(f(x)\) is never zero.
When \(x=0\), we have \(y=0\). This is the only intersection with the axes.
As \(x \rightarrow \infty\), \(y\) behaves like \(\dfrac{1}{x}\) and so \(y \rightarrow 0\).
Looking for stationary points, we can differentiate using the quotient rule and set \(\dfrac{dy}{dx}=0\): \[\begin{align*} \frac{dy}{dx} = \frac{6(x^2+x+1)-6x(2x+1)}{(x^2+x+1)^2} &= 0\\ \iff \qquad -6x^2+6 &=0,\\ \end{align*}\]and this holds if and only if \(x=\pm 1\).
When \(x = 1\), we have \(y = 2\), and when \(x = -1\) we have \(y = -6\) (which are precisely \(\dfrac{a}{3}\) and \(-a\) as expected).
By drawing on the same diagram a certain straight line show that the equation \(x^3-1=6x\) has three real roots, two negative and one positive.
Note that \[x^3-1=(x-1)(x^2+x+1)\] so the equation \(x^3-1=6x\) is equivalent to \[x-1= \frac{6x}{x^2+x+1}\] (we saw above that \(x^2 + x + 1 \neq 0\) for real \(x\)).
Thus, adding the line \(y=x-1\),
we can see that \(x^3 - 1 = 6x\) has two negative roots and one positive.
