Find all possible values of $\frac{1}{{x^2 - x - 1}}$

Question

Question: Find all possible values of $$\frac{1}{{x^2 - x - 1}}$$

My solution-> $\frac{1}{{x^2 - x - 1}}=\frac{1}{{(x-\frac{1}{2})^2-\frac{5}{4}}}$
=$(x-\frac{1}{2})^2 \geq 0 \quad \forall x \in \mathbb{R}$
=$(x-\frac{1}{2})^2 - \frac{5}{4} \geq -\frac{5}{4}$
=$- \frac{5}{4} \leq (x-\frac{1}{2})^2 - \frac{5}{4} < \infty$
=$- \frac{5}{4} \geq \frac{1}{{(x-\frac{1}{2})^2 - \frac{5}{4}}} > 0$
=$[-\frac{5}{4}, 0) $

book showing the answer to be $$\frac{1}{{\left(x-\frac{1}{2} \right)^2 -\frac{5}{4}}} \in \left(-\infty, -\frac{4}{5} \right] \cup (0, +\infty)$$ Please provide the whole process.

Answer 1

=$- \frac{5}{4} \leq (x-\frac{1}{2})^2 - \frac{5}{4} < \infty$
=$- \frac{5}{4} \geq \frac{1}{{(x-\frac{1}{2})^2 - \frac{5}{4}}} > 0$

$\dfrac{1}{\infty} = 0$, but $\dfrac{1}{-\dfrac{5}{4}} = -\dfrac{5}{4}?$

Also, reconsider the statement $- \dfrac{5}{4} \geq \dfrac{1}{{(x-\frac{1}{2})^2 - \frac{5}{4}}} > 0$. What happens when $(x-\dfrac{1}{2})^2 - \dfrac{5}{4} =0?$

When dealing with the range of complicated functions, always switch to an easier perspective of domain. Ie.

$$\text{Let } y =\dfrac{1}{{(x-\frac{1}{2})^2 - \frac{5}{4}}}, \text{ then the inverse is } \\ x = \sqrt{\dfrac{1}{y} + \dfrac{5}{4}} + \dfrac{1}{2}$$

Now, asking to

find all possible values of $\dfrac{1}{{(x-\frac{1}{2})^2 - \frac{5}{4}}}$

is the same as asking

what is the domain of the function $\sqrt{\dfrac{1}{y} + \dfrac{5}{4}} + \dfrac{1}{2}$, or simply, at what values of $y$ does $x$ exist.

Now, start off seeing the restrictions to the domain. Ie. $\dfrac{1}{y} + \dfrac{5}{4} \geq 0$.

Answer 2

I'm sure this isn't sufficiently rigorous, but here's my attempt:

Let $p(x) = x^2-x-1$. Being the sum of finitely many nonnegative integer powers of $x$, this is clearly a polynomial and as such is defined for all $x\in \mathbb{R}$. Via the quadratic formula its roots are $r_{1}=\frac{1+\sqrt{5}}{2}$ and $r_{2}=\frac{1-\sqrt{5}}{2}$. Furthermore, since the coefficient on $x^2$ is positive we know that the parabola representing this function opens upwards, i.e. $p(x)\lt0$ for $x\in(\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$, with $p(x) \ge 0$ otherwise. Via calculus, the minimum of $p(x)$ occurs when $x=\frac{1}{2}$. The value of $p(\frac{1}{2})$ works out to be $\frac{-5}{4}$.

Now consider $f(x) = \frac{1}{x^2-x-1}$. You can, in a somewhat dodgy sense (I believe it to be), consider this as being equal to $\frac{p(x)}{(p(x))^2}$, at least as long as $p(x)\neq 0$. Because $(p(x))^2 \gt 0$ for all $p(x)\in (-\infty,0)\cup(0,\infty)$, if $p(x) \lt 0$ or $p(x) \gt0$, then $f(x) \lt0$ or $f(x)\gt0$. In other words, $f(x)$ and $p(x)$ have the same sign.

That being the case, it is clear that $|f(x)| = \frac{1}{|p(x)|}$. To find the range of $f(x)$ therefore requires you to find the range of $p(x)$, keeping in mind this equality and the fact that they will always have the same sign. For $x\in(-\infty,\frac{1-\sqrt{5}}{2})$, $p(x)\in(0,\infty)$, and likewise for $x\in(\frac{1+\sqrt{5}}{2},\infty)$, which means that $f(x)\in(0,\infty)$. This accounts for the interval on the RHS of the union in the given answer.

Now consider $x\in(\frac{1-\sqrt{5}}{2},\frac{1}{2}]$. Recall that $p(x)\lt0$ over this interval and that $p(\frac{1}{2})=\frac{-5}{4}$. In other words, for all $x$ in this interval, $p(x)\in[\frac{-5}{4},0)$. Since the value of $f(x)$ is equal to $\frac{1}{p(x)}$ for all values of $x$ such that $p(x)\neq0$, we can conclude that for values of $x$ in the above interval, $f(x)\in(-\infty,\frac{-4}{5}]$.

Put these intervals together and you get the correct answer, $f(x)\in(-\infty,\frac{-4}{5}]\cup(0,\infty)$.

Notes: I did play a bit fast and loose with rigour I think, but I'm pretty confident in my reasoning. You might be wondering, what about the "missing" $x$; that is, what about $x\in[\frac{1}{2},\frac{1+\sqrt{5}}{2})$? You can work it out if you like (and I encourage you to) and see for yourself that you end up at the same conclusion you get from analysing $p(x)$ and $f(x)$ for $x\in(\frac{1-\sqrt{5}}{2},\frac{1}{2}]$. This is because $p(x)$ (and therefore $f(x)$) is symmetric about the line $x=\frac{1}{2}$.

I hope this answer helps. If you or anybody else spots a flaw in my work (or have suggestions as to where I could improve), I'd be appreciative if you could let me know. I'm pretty new at this in earnest, you see. In the meantime: peace!

Answer 3

Let $$y = \frac{1}{x^2 - x - 1}$$ Then \begin{align*} yx^2 - yx - y & = 1\\ yx^2 - yx - y - 1 & = 0 \end{align*} which is a quadratic equation in $x$ unless $y = 0$.

If $y = 0$, then $$\frac{1}{x^2 - x - 1} = 0$$ Multiplying both sides of the equation by $x^2 - x - 1$ yields $1 = 0$, which is a contradiction. Thus, $0$ is not in the range.

Hence, a real number $y \neq 0$ is in the range of the function if and only if there exists a real number $x$ in the domain of the function that satisfies the quadratic equation $yx^2 - yx - y - 1 = 0$. A quadratic equation $ax^2 + bx + c = 0$ with real coefficients $a, b, c$, where $a \neq 0$, has real roots if and only if its discriminant $\Delta = b^2 - 4ac \geq 0$.

\begin{align*} \Delta & \geq 0\\ (-y)^2 - 4y(-y - 1) & \geq 0\\ y^2 + 4y(y + 1) & \geq 0\\ y^2 + 4y^2 + 4y & \geq 0\\ 5y^2 + 4y & \geq 0\\ y(5y + 4) & \geq 0 \end{align*} Equality holds if \begin{align*} y & = 0 & \text{or} & & 5y + 4 & = 0\\ & & & & 5y & = -4\\ & & & & y & = -\frac{4}{5} \end{align*}

$y(5y + 4) > 0$ if $y$ and $5y + 4$ have the same sign, which occurs when $y > 0$ or $y < -4/5$. Hence, $\Delta \geq 0$ if $y \geq 0$ or $y \leq -4/5$. However, we showed above that $y = 0$ is not in the range. Therefore, $y$ is in the range if $y > 0$ or $y < -4/5$. Hence, the range of the function is $$\left(-\infty, -\frac{4}{5}\right] \cup (0, \infty)$$