Problem 6 - Essential Calculus

Question

Find the two \(x\) -intercepts of the function \(f\) and show that \(f^{\prime}(x)=0\) at some point between the two \(x\) -intercepts. $$ f(x)=x(x-3) $$

Step-by-Step Solution

Answer

The x-intercepts are at \(x = 0\) and \(x = 3\). \(f'(x) = 0\) for \(x = 1.5\), which is indeed located between the x-intercepts.

Step 1: Find the x-intercepts

Set the function \(f(x) = x(x - 3)\) equal to 0 and solve for x: \n\n\(f(x) = 0\)\n\(x(x - 3) = 0\)\nThis equation will be true if either \(x = 0\) or \(x = 3\). So the x-intercepts are 0 and 3.

Step 2: Find the derivative

The derivative of the function \(f(x) = x(x - 3)\) is given by applying the product rule for differentiation: \n\n\(f^{\prime}(x) = 1*(x-3) + x*1\)\nThis simplifies to: \n\n\(f^{\prime}(x) = x - 3 + x = 2x - 3\)

Step 3: Set the derivative equal to 0 and solve for x

To find where the derivative equals zero, set \(f^{\prime}(x) = 0\) and solve for \(x\): \n\n\(2x - 3 = 0\)\n\(2x = 3\)\n\(x = 3/2\) or \(x = 1.5\)\nSince 1.5 is between the x-intercepts (0 and 3), there's indeed a point on the graph of the function where the slope of the tangent is 0.

Key Concepts

Understanding X-intercepts

X-intercepts represent the points where a graph crosses the x-axis. These are places where the value of the function is zero. To find the x-intercepts, set the function equal to zero and solve for the variable. For the function \(f(x) = x(x - 3)\), we set \(f(x) = 0\), which gives us two solutions: \(x = 0\) and \(x = 3\).
This means the graph of the function will touch or cut the x-axis at these points. Identifying x-intercepts is crucial for understanding the function's behavior and roots.

Grasping the Derivative

A derivative represents the rate of change of a function. It's like finding the "speed" at which the function's value is moving as the input changes. For example, if you are analyzing the function \(f(x) = x(x - 3)\), the derivative shows how fast \(f(x)\) is changing with respect to \(x\). In our case, the derivative \(f^{\prime}(x)\) is calculated using differentiation rules to find \(2x - 3\).
Derivatives help in understanding the slope of the tangent line to the graph at any given point, which is essential in many applications such as physics and engineering.

The Product Rule Demystified

The product rule is a fundamental tool in calculus used to differentiate functions that are products of two simpler functions. It's crucial when dealing with functions like \(f(x) = x(x - 3)\). The product rule states that the derivative of a product \(u(x)v(x)\) is \(u^{\prime}(x)v(x) + u(x)v^{\prime}(x)\).
Applying this to our function means finding the derivative of \(x\) and \(x - 3\) separately and then combining them. Hence, we find \(f^{\prime}(x) = 1\cdot(x - 3) + x\cdot1 = 2x - 3\). This rule is essential for more advanced calculus problems that involve product of functions.

Understanding Tangents and Their Application

A tangent to a curve is a line that touches the curve at a single point without crossing it. The slope of the tangent line at any point on a curve is given by the derivative at that point. For the function \(f(x) = x(x - 3)\), the derivative \(f^{\prime}(x) = 2x - 3\) determines the slope of the tangent.
When \(f^{\prime}(x) = 0\), the tangent is horizontal, indicating a local maximum or minimum point on the curve. In our problem, this occurs at \(x = 1.5\), which lies between the x-intercepts \(0\) and \(3\). This is vital for understanding optimization and curve sketching.

Principles of Differentiation

Differentiation is the process of calculating a derivative and is a foundational concept in calculus. It helps in understanding how functions behave, grow, and change. By using rules like the power rule, product rule, and chain rule, we can find derivatives for various types of functions.
In the context of \(f(x) = x(x - 3)\), differentiation helps in understanding the variation of the function and in solving for critical points, like where the function has horizontal tangents. Mastery of differentiation is essential for solving not just mathematical problems, but also real-world problems in science and economics.