# How to Find the Number of Roots by Stationary Points and Deduce for an Inverse Function Explained Quickly

The stationary points can be found by solving the derivatives to zero. Some cubic equations have one root if the stationary points are in the same sign, such as two stationary points are either above the x-axis or below the x-axis. In these cases, the cubic graph cuts the x-axis only once, so the cubic equation has one root.

If the functions do not have stationary points, then they are either monotonic increasing or decreasing, which results in one-to-one functions. This is enough to tell that the functions have inverse functions.

Now, we will see these two theories through the following examples. Let’s get into it!

Let $f(x) = Ax^3 – Ax + 1$, where $A \gt 0$.

## Part 1

Find the x-coordinates of the stationary points of $f(x)$.

\begin{align} \displaystyle f(x) &= Ax^3 – Ax + 1 \\ f^{\prime}(x) &= 3Ax^2 – A \\ 3Ax^2 – A &= 0 &\text{stationary points occur when } f(x) = 0 \\ 3Ax^2 &= A \\ 3x^2 &= 1 \\ x^2 &= \frac{1}{3} \\ \therefore x &= \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3} \end{align}

## Part 2

Find $A$ so that $f(x)$ has exactly one root.

Since $f(x)$ has stationary points at $\displaystyle x = \pm \frac{\sqrt{3}}{3}$, and $A \gt 0$, and the $y$-intercept $= 1$, then if there is only one root, the graph of $f(x)$ is:

\begin{align} \displaystyle f \left(\frac{\sqrt{3}}{3} \right) &\gt 0 \cdots (1) &\text{The minimum stationary point is above the } x \text{-axis} \\ f \left(\frac{\sqrt{3}}{3} \right) &= A \left(\frac{\sqrt{3}}{3} \right)^3 – A \left(\frac{\sqrt{3}}{3} \right) +1 \\ &= \frac{3\sqrt{3}A}{27} – \frac{\sqrt{3}A}{3} + 1 \\ &= \frac{\sqrt{3}A}{9} – \frac{\sqrt{3}A}{3} + 1 \\ &= \frac{\sqrt{3}A – 3\sqrt{3}A}{9} + 1 \\ &= \frac{-2\sqrt{3}A}{9} + 1 \\ \frac{-2\sqrt{3}A}{9} + 1 &\gt 0 \cdots (1) \\ \frac{-2\sqrt{3}A}{9} &\gt -1 \\ A &\lt \frac{9}{2\sqrt{3}} \\ A &\lt \frac{9}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \\ \therefore A &\lt \frac{3\sqrt{3}}{2} \end{align}

## Part 3

Deduce that $f(x)$ does not have a root in the interval $-1 \le x \le$ when $\displaystyle 0 \lt A \lt \displaystyle \frac{3 \sqrt{3}}{2}$.

Now if $f(-1) = 1$, and there is only one root when $\displaystyle A \lt \frac{3 \sqrt{3}}{2}$, then the root must occur when $x \lt -1$. Therefore there is no root in the interval $-1 \le x \le 1$.

## Part 4

Let $g(\theta) = 2 \cos \theta + \tan \theta$, where $\displaystyle -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$. Show that $g(\theta)$ does not have any stationary points.

\begin{align} \displaystyle g^{\prime} (\theta) &= -2 \sin \theta + \sec^2 \theta \\ -2 \sin \theta + \sec^2 \theta &= 0 \\ -2 \sin \theta + \frac{1}{\cos^2 \theta} &= 0 \\ -2 \sin \theta + \frac{1}{1 – \sin^2 \theta} &= 0 \\ -2 \sin \theta (1 – \sin^2 \theta) + 1 &= 0 \\ -2 \sin \theta + 2 \sin^3 \theta +1 &= 0 \\ 2 \sin^3 \theta – 2 \sin \theta +1 &= 0 \\ 2x^3 – 2x + 1 &= 0 &\text{by letting } x = \sin \theta \\ Ax^3 – Ax +1 &= 0 \leadsto A = 2 \\ \therefore 0 \lt A \lt \frac{3\sqrt{3}}{2} \\ -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2} \\ -1 \lt \sin \theta \lt 1 \end{align}

Therefore g^{\prime}(\theta) \ne 0 and $g(\theta)$ has no stationary points for $\displaystyle -\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}$.

## Part 5

Deduce that $g(\theta)$ has an inverse function.

Since $g(\theta)$ has no stationary points, then it is either monotonic increasing or decreasing, and it will have an inverse function. 