Problem 108 - Calculus Volume 1

Question

For the following exercises, find the critical points in the domains of the following functions. $$y=4 x^{3}-3 x$$

Step-by-Step Solution

Answer

The critical points are at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

Step 1: Differentiate the Function

To find the critical points, the first step is to take the derivative of the function. Given the function \( y = 4x^3 - 3x \), differentiate it with respect to \( x \). The derivative is given by \( \frac{d}{dx}(4x^3 - 3x) = 12x^2 - 3 \).

Step 2: Set the Derivative Equal to Zero

Critical points occur where the derivative is zero or undefined. Set the derivative \( 12x^2 - 3 \) equal to zero and solve for \( x \):\[12x^2 - 3 = 0\]\[12x^2 = 3\]\[x^2 = \frac{1}{4}\]

Step 3: Solve for Critical Points

Solve the equation \( x^2 = \frac{1}{4} \) to find the points where the function reaches critical points:\[x = \pm \frac{1}{2}\]Thus, the critical points occur at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \).

Key Concepts

Derivative

Derivatives play a crucial role in calculus, especially when it comes to understanding the behavior of functions. When we talk about derivatives, we refer to the process of finding the rate at which a function changes at a particular point. To understand this, we differentiate the function with respect to its variable, often using well-established rules like the power rule or the chain rule.
\[\frac{d}{dx}(4x^3 - 3x) = 12x^2 - 3\]

• Interpretation: The derivative \(12x^2 - 3\) gives us the slope of the tangents to the curve \(y = 4x^3 - 3x\).
• Slope significance: A positive slope implies an increasing function, a negative slope implies a decreasing function, and a slope of zero indicates potential critical points.

By finding the derivative, we get a formula that describes how the function behaves. It's essential for identifying areas where the function's direction changes, known as critical points.

Solve Equation

Once we have the derivative, the next step is solving the equation where we set the derivative equal to zero. This is crucial because critical points often occur where the derivative is zero.
\[12x^2 - 3 = 0\]

• Step-by-step solution: First, add 3 to both sides to obtain \(12x^2 = 3\).
• Next, divide by 12 to isolate \(x^2\), leading to \(x^2 = \frac{1}{4}\).
• Finally, take the square root of both sides, giving \(x = \pm \frac{1}{2}\).

This process reveals the potential values of \(x\) where the function could change its behavior. Critical points are significant because they often correspond to local minima, maxima, or points of inflection.

Function Analysis

Analyzing a function involves looking at its critical points and understanding its overall behavior. When we find critical points, such as those determined by setting the derivative to zero, we gather insights into where the function might change direction.

• In our example, the critical points \(x = \frac{1}{2}\) and \(x = -\frac{1}{2}\) tell us where the slope of the tangent to the curve is zero.
• These points are crucial in identifying where the function \(y = 4x^3 - 3x\) could potentially have a local maximum or minimum.

By using the first derivative test or the second derivative test, you can determine the nature of these critical points. For instance, the function is decreasing when the derivative is negative, increasing when the derivative is positive, and flat when the derivative is zero.
This analysis empowers you to sketch the graph accurately and foresee how the function behaves across its domain.