Problem 51 - University Calculus: Early Transcendentals

Question

Determine all critical points for each function. $$y=x^{2}-32 \sqrt{x}$$

Step-by-Step Solution

Answer

The critical point is at \(x = 4\).

Step 1: Identify the Function

We start with the function given by \[ y = x^2 - 32 \sqrt{x}\].

Step 2: Find the First Derivative

To find the critical points, we need to determine where the derivative of the function is zero or undefined. First, differentiate the function:\[ \frac{dy}{dx} = 2x - \frac{32}{2\sqrt{x}} = 2x - \frac{16}{\sqrt{x}}\].

Step 3: Simplify the First Derivative

The simplified form of the derivative can be obtained by rewriting it with a common denominator:\[ \frac{dy}{dx} = \frac{2x\sqrt{x} - 16}{\sqrt{x}}\].

Step 4: Set the First Derivative to Zero

Next, set the first derivative equal to zero to find critical points:\[ \frac{2x\sqrt{x} - 16}{\sqrt{x}} = 0\].Multiply through by \(\sqrt{x}\) to clear the denominator:\[ 2x\sqrt{x} = 16 \].

Step 5: Solve for x

Solve the equation \(2x\sqrt{x} = 16\). Divide both sides by 2:\[ x\sqrt{x} = 8\].Set \(u = \sqrt{x}\), then \(x = u^2\), and substitute to get:\[ u^2u = 8 \Rightarrow u^3 = 8\].Solving for \(u\), we get \(u = 2\). Since \(u = \sqrt{x}\), we have \(\sqrt{x} = 2\), leading to \(x = 4\).

Step 6: Check for Undefined Points

Consider where the derivative is undefined, which means the denominator \(\sqrt{x}\) cannot be zero, implying \(x eq 0\).

Step 7: Conclusion: List the Critical Points

The critical point of the function is \(x = 4\). Since \(x = 0\) would make the derivative undefined and is not in the domain of the original function (as you cannot find the square root of a negative number), it is not considered a critical point.

Key Concepts

First Derivative

The first derivative of a function is a crucial tool in calculus, as it helps us identify various characteristics of the original function, such as its slope at any given point. For the function \[ y = x^2 - 32 \sqrt{x} \],we begin by differentiating it with respect to \(x\). This process gives us:

• The derivative of \(x^2\) is \(2x\).
• The derivative of \(-32\sqrt{x}\) requires a bit more work since \(\sqrt{x}\) is equivalent to \(x^{1/2}\). Therefore, its derivative is \(\frac{-32}{2\sqrt{x}}\), which simplifies to \(-\frac{16}{\sqrt{x}}\).

By putting them together, we find that the first derivative is \[ \frac{dy}{dx} = 2x - \frac{16}{\sqrt{x}} \].This derivative captures how the function \(y\) changes with respect to small changes in \(x\).

Function Differentiation

Function differentiation involves the process of finding the derivative of a function. In our specific case, \[ y = x^2 - 32 \sqrt{x} \],we apply rules of differentiation to break down the function:

• The power rule: For any term \(ax^n\), the derivative is \(anx^{n-1}\).
• The chain rule: If a function is composed of another function, like \(32\sqrt{x} = 32x^{1/2}\), we derive it step-by-step by applying the chain rule.

Differentiating may often involve combining rules, as seen with the term involving the square root. Understanding differentiation helps navigate the complexities of various mathematical functions, allowing for more accurate and meaningful analysis.

Critical Point Analysis

Critical points in a function occur where its first derivative is zero or undefined. These points help us understand where the function’s graph changes direction. After finding the first derivative \[ \frac{dy}{dx} = \frac{2x\sqrt{x} - 16}{\sqrt{x}} \],we analyze it.

• To find critical points, set the derivative to zero: \(\frac{2x\sqrt{x} - 16}{\sqrt{x}} = 0\).
• Clear the denominator by multiplying through by \(\sqrt{x}\), resulting in \(2x\sqrt{x} = 16\).
• Solving it gives us \(x = 4\), which indicates a critical point because at this value, the slope of the tangent to the function \(y\) is zero.

Understanding critical point analysis is vital for discovering local maxima and minima in functions, which can provide insight into function behavior across intervals.

Derivative Simplification

Once the first derivative is determined, simplifying it is an essential next step to make further analysis more manageable. The derivative for our example started as \[ \frac{dy}{dx} = 2x - \frac{16}{\sqrt{x}} \].To simplify:

• Express the terms with a common denominator: \(\frac{2x\sqrt{x}}{\sqrt{x}} - \frac{16}{\sqrt{x}}\).
• Combine these terms under the same denominator: \(\frac{2x\sqrt{x} - 16}{\sqrt{x}}\).

This simplification helps us set the derivative equal to zero more easily to find critical points. Simplifying derivatives is key in calculus, as it aids in uncovering the relationships between variables more clearly, making solving equations less cumbersome.