In this article, we will solve an inequality using pure algebra instead of the conventional graph sketching method.
Given $\displaystyle f(x) = \frac{x^2 + 1}{(x+1)(x-7)}$, find the set of values of $x$ that satisfy $f(x) > \dfrac{1}{f(x)}$.
We first note that $f(x) > 0 \iff x<-1 \vee x> 7$ and $f(x) < 0 \iff -1 < x < 7$ by sign test.
Case #1: $f(x) > 0$
\[\begin{aligned}
& f(x) > \frac{1}{f(x)} \\
\implies & (f(x))^2 - 1 > 0 \\
\iff & (f(x)-1)(f(x)+1) > 0 \\
\end{aligned}\]
From the inequality above, $f(x) < -1$ (reject since $f(x) > 0$) and $f(x) > 1$.
When $f(x) > 1$,
\[\begin{aligned}
& \frac{x^2 + 1}{(x+1)(x-7)} > 1 \\
\iff & \frac{6x + 8}{(x+1)(x-7)} > 0 \\
\iff & \frac{x + \frac{4}{3}}{(x+1)(x-7)} > 0 \\
\end{aligned}\]
By sign test, $-\dfrac{4}{3} < x < -1$ and $x > 7$.
Case #2: $f(x) < 0$
\[\begin{aligned}
& f(x) > \frac{1}{f(x)} \\
\implies & (f(x))^2 - 1 < 0 \\
\iff & (f(x)-1)(f(x)+1) < 0 \\
\end{aligned}\]
From the inequality above, $-1 < f(x)< 0$ .
This gives
\(-1 < \frac{x^2 + 1}{(x+1)(x-7)} < 0\)
This case works for $-1 < x < 7$, so $(x+1)(x-7) < 0$.
We split the inequality into two parts again.
\[\begin{aligned}
& \frac{x^2 + 1}{(x+1)(x-7)} < 0 \\
\iff & x^2 + 1 > 0 \\
\iff & x^2 > -1 \\
\implies & x \in \mathbb{R} \\
\end{aligned}\]
\[\begin{aligned}
& \frac{x^2 + 1}{(x+1)(x-7)} > -1 \\
\implies & x^2 + 1 < 6x +7 -x^2 \\
\iff & x^2 -3x -3 < 0 \\
\implies & \frac{3-\sqrt{21}}{2} < x < \frac{3+\sqrt{21}}{2} \\
\end{aligned}\]
From case #2, we conclude $\dfrac{3-\sqrt{21}}{2} < x < \dfrac{3+\sqrt{21}}{2}$.
Thus, \(x \in \boxed{\left(-\frac{4}{3},1\right) \cup \left(\frac{3-\sqrt{21}}{2},\frac{3+\sqrt{21}}{2}\right) \cup (7,\infty)}.\)
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