Problem 11 - Thomas Calculus in SI Units

Question

Explain why the following four statements ask for the same information: i) Find the roots of \(f(x)=x^{3}-3 x-1\). ii) Find the \(x\) -coordinates of the intersections of the curve \(y=x^{3}\) with the line \(y=3 x+1\). iii) Find the \(x\) -coordinates of the points where the curve \(y=x^{3}-3 x\) crosses the horizontal line \(y=1\). iv) Find the values of \(x\) where the derivative of \(g(x)=\) \((1 / 4) x^{4}-(3 / 2) x^{2}-x+5\) equals zero.

Step-by-Step Solution

Answer

All four statements lead to solving the same equation: \(x^3 - 3x - 1 = 0\).

Step 1: Understanding Statement i

The statement asks us to find the roots of the function \(f(x) = x^3 - 3x - 1\). Finding the roots of a function means finding the values of \(x\) for which \(f(x) = 0\). Thus, we need to solve the equation \(x^3 - 3x - 1 = 0\).

Step 2: Analyzing Statement ii

This statement concerns finding where a cubic curve \(y = x^3\) intersects a line \(y = 3x + 1\). To find the \(x\)-coordinates of the intersections, we equate the two expressions for \(y\): \(x^3 = 3x + 1\). Simplifying this results in the equation \(x^3 - 3x - 1 = 0\), which is the same as in statement i.

Step 3: Examining Statement iii

This statement asks for the \(x\)-coordinates where the graph of \(y = x^3 - 3x\) crosses the line \(y = 1\). To find the points of intersection, we equate these expressions: \(x^3 - 3x = 1\). Simplifying, we have \(x^3 - 3x - 1 = 0\), which is again the same equation as in statement i.

Step 4: Understanding Statement iv

This statement involves finding values of \(x\) where the derivative of \(g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5\) equals zero. First, we find the derivative: \(g'(x) = x^3 - 3x - 1\). Setting \(g'(x) = 0\) leads to \(x^3 - 3x - 1 = 0\), matching the equation in statement i.

Key Concepts

Intersection of Curves

When we talk about the intersection of curves, we're usually referring to points where two different graphical representations share a common space. In this context, it relates to when curves intersect with each other on an x-y plane. Imagine a graph where you have two functions:

• Curve 1: \( y = x^3 \)
• Line: \( y = 3x + 1 \)

To find their points of intersection, you need to find values of \( x \) where both equations give the same \( y \)-value. This means setting them equal to one another: \( x^3 = 3x + 1 \), simplifying to \( x^3 - 3x - 1 = 0 \).
You solve this cubic equation to find the x-values where these two curves intersect.
Intersection points are quite important because they are often where something significant happens, like where two forces balance out or two trends meet. Recognizing intersections is crucial in problem-solving as they often reveal solutions to seemingly complex problems. Each intersection corresponds to a root of the polynomial that results from equating the equations, tying nicely into finding the roots of polynomials, as seen in the exercise.

Derivative of Functions

The derivative of a function gives us a lot of valuable information about the behavior of the function itself. When you take the derivative, you're essentially finding the slope of the tangent to the function at any point. This is the rate at which the function changes.
For the function given, \( g(x) = \frac{1}{4}x^4 - \frac{3}{2}x^2 - x + 5 \), the derivative gives: \[ g'(x) = x^3 - 3x - 1 \]By setting the derivative equal to zero, \( g'(x) = 0 \), we're finding points where the slope of the tangent is horizontal. These are the critical points where the function could have a maximum, minimum, or a point of inflection.
In the specific context of this problem, setting the derivative to zero led to the equation \( x^3 - 3x - 1 = 0 \), the same as finding the roots of the polynomial in the exercise. This intersection between differentiating and finding roots is a powerful concept in calculus, highlighting how derivatives help us identify essential characteristics of functions.

Cubic Equations

Cubic equations are those of the form \( ax^3 + bx^2 + cx + d = 0 \). Solving them means finding the values of \( x \) that satisfy the equation. These are called the roots of the polynomial.
In the exercise, each statement essentially required solving the cubic equation \( x^3 - 3x - 1 = 0 \). Cubic equations are interesting because, unlike quadratics, which always have a closed-form solution using radicals, they need more complex methods or numerical approximations for their solutions.
One approach to solving a cubic equation is factoring. In some cases, you can find one root by inspection or trial and error, which helps to factor the cubic into a product of a linear and a quadratic equation.
Alternatively, methods such as the Rational Root Theorem or synthetic division can also be helpful. Often, numerical methods or tools might come in handy to approximate the roots when they cannot be easily factored.
Understanding cubic equations is crucial as they appear in various real-world applications, from physics to engineering, offering vast insights into the nature of polynomials and their roots.