# Master Root-Finding by Stationary Points in 5 Simple Steps

# Finding the Number of Roots in a Cubic Equation Using Stationary Points

Finding the number of roots in a cubic equation can be challenging, but by analyzing the stationary points, you can quickly determine the number of roots without actually solving the equation. In this tutorial, we’ll guide you through a simple 5-step process to master root-finding using stationary points.

## Understanding Stationary Points

Stationary points are points on a graph where the derivative of the function is equal to zero. In other words, they are points where the slope of the tangent line is zero. Moreover, stationary points can be classified into three types: local minimum, local maximum, and point of inflection.

## Finding the Derivative

To find the stationary points, you first need to find the derivative of the cubic function. The derivative of a cubic function is a quadratic function, which can be found using the power rule. For instance, if the cubic function is \( f(x) = x^3 + bx^2 + cx + d \), then the derivative is \( f'(x) = 3x^2 + 2bx + c \).

## Solving the Derivative

Once you have the derivative, set it equal to zero and solve for \( x \). The solutions to this equation are the \(x\)-coordinates of the stationary points. Depending on the values of \( b \) and \( c \), the derivative may have zero, one, or two distinct solutions.

## Analysing the Stationary Points

### No Real Solutions

If the derivative has no real solutions, then the cubic function has no stationary points. In this case, the function is either monotonically increasing or decreasing, resulting in a one-to-one function with a single root.

### One Solution

On the other hand, if the derivative has one solution, then the cubic function has one stationary point, which is a point of inflection. In this case, the function may have one or three roots, depending on the y-coordinate of the stationary point.

### Two Distinct Solutions

However, if the derivative has two distinct solutions, then the cubic function has two stationary points. If both stationary points are above the x-axis or both are below the x-axis, then the cubic function has only one root. This is because the function’s graph will only cross the x-axis once, regardless of the shape of the graph between the stationary points.

## Deducing the Number of Roots

By analyzing the stationary points, you can deduce the number of roots without actually solving the cubic equation. When there are no stationary points, the function has one root. In contrast, when there is one stationary point, the function may have one or three roots. Furthermore, when there are two stationary points on the same side of the x-axis, the function has one root.

## Inverse Functions and Monotonicity

In addition to root-finding, stationary points can also help determine if a function has an inverse. If a function has no stationary points and is either monotonically increasing or decreasing, then it is a one-to-one function and has an inverse function. One-to-one functions have the property that each x-value corresponds to a unique y-value, and vice versa.

Mastering the concept of stationary points allows you to quickly determine the number of roots in a cubic equation and deduce the existence of inverse functions. This powerful tool can save you time and effort in analyzing complex functions and equations. Therefore, remember to practice these steps and apply them to various cubic functions to reinforce your understanding. With these skills in your mathematical toolkit, you’ll be well-equipped to tackle a wide range of problems in algebra, calculus, and beyond.

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

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