Problem 126 - Calculus Volume 1

Question

Find the local and/or absolute maxima for the functions over the specified domain. \(y=\sqrt{x}-\sqrt{x^{3}}\) over [0,4]

Step-by-Step Solution

Answer

The absolute maximum is at \(x = \frac{1}{3}\) with \(y = \frac{2\sqrt{3}}{9}\).

Step 1: Define the function

Given the function is \( y = \sqrt{x} - \sqrt{x^3} \). We need to find the local and absolute maxima for this function over the interval \([0, 4]\).

Step 2: Find the first derivative

To find critical points, we must first find the derivative of \( y \). Using the power rule, the derivative is:\[\frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{3x^{2}}{2\sqrt{x^3}} \]Simplifying this, we can write the derivative as:\[\frac{dy}{dx} = \frac{1 - 3x}{2\sqrt{x}} \]

Step 3: Determine critical points

Set the first derivative equal to zero to find the critical points:\[\frac{1 - 3x}{2\sqrt{x}} = 0 \]Solving \(1 - 3x = 0\) gives \(x = \frac{1}{3}\). Check the domain \([0, 4]\) to ensure this critical point is within the interval. This critical point, \(x = \frac{1}{3}\), is valid.

Step 4: Evaluate the function at critical points and endpoints

Evaluate \( y = \sqrt{x} - \sqrt{x^3} \) at the critical point and the endpoints of the domain:For \(x = 0\):\[y = \sqrt{0} - \sqrt{0^3} = 0\]For \(x = \frac{1}{3}\):\[y = \sqrt{\frac{1}{3}} - \sqrt{(\frac{1}{3})^3} = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9}\]For \(x = 4\):\[y = \sqrt{4} - \sqrt{4^3} = 2 - 8 = -6\]

Step 5: Compare the values

Compare the values obtained:- At \(x = 0\), \( y = 0 \)- At \(x = \frac{1}{3}\), \( y = \frac{2\sqrt{3}}{9} \)- At \(x = 4\), \( y = -6 \)Since \(y = \frac{2\sqrt{3}}{9} \approx 0.385\) is the highest value, this is the local and absolute maximum on the domain \([0, 4]\).

Key Concepts

Understanding the First Derivative

The first derivative of a function is a fundamental concept in calculus that provides valuable information about the function's slope at any given point. Imagine you're hiking; the first derivative is like the slope of the ground you're walking on.
If the slope is positive, you're going uphill, and if it's negative, you're going downhill. When working with the function \( y = \sqrt{x} - \sqrt{x^3} \), finding its first derivative helps us understand how the function behaves over its domain. We used the power rule to find:

• \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} - \frac{3x^{2}}{2\sqrt{x^3}} \), which simplifies to \( \frac{dy}{dx} = \frac{1 - 3x}{2\sqrt{x}} \).

By analyzing this derivative, we can identify changes in direction of the function, helping us to pinpoint critical points where potential maxima or minima might occur.

Locating Critical Points

Critical points occur where the first derivative of a function is zero or undefined. They are key in identifying points where the function could change from increasing to decreasing, or vice versa.
For our function, we set the first derivative \( \frac{1 - 3x}{2\sqrt{x}} \) equal to zero to find its critical points:

• Solve \( 1 - 3x = 0 \), which provides \( x = \frac{1}{3} \).

Critical points are important to check within the specified domain—in this case, \([0, 4]\). Thus, \( x = \frac{1}{3} \) is a valid critical point on this interval. This investigation lays the groundwork to understand where local maxima or minima could occur.

Defining a Local Maximum

A local maximum of a function is a point where the function reaches a peak within a neighborhood, without being the absolute highest point of the entire domain. It's like reaching the top of a hill.
To determine if a critical point is a local maximum, evaluate the function at this point and its behaviors nearby. When we evaluated our function \( y = \sqrt{x} - \sqrt{x^3} \) at \( x = \frac{1}{3} \):

• \( y = \frac{2\sqrt{3}}{9} \) was the result, approximately 0.385.

In the context of the problem, this value turned out to be the highest among the points evaluated, suggesting not just a local peak, but also bearing the distinctive title of absolute maximum for this particular domain.

Recognizing an Absolute Maximum

An absolute maximum is the largest value a function achieves over its entire domain. It's the highest peak, like the summit of a mountain. To discover this maximum, we check both critical points and the boundaries of the domain.
In our example, we evaluated the function \( y = \sqrt{x} - \sqrt{x^3} \) at the endpoints \( x = 0 \) and \( x = 4 \) alongside the critical point:

• For \( x = 0 \), \( y = 0 \)
• For \( x = \frac{1}{3} \), \( y = \frac{2\sqrt{3}}{9} \)
• For \( x = 4 \), \( y = -6 \)

Among these values, \( \frac{2\sqrt{3}}{9} \) was the greatest, making it the absolute maximum over the interval \([0, 4]\). Finding such maxima is crucial in understanding the full behavior of the function over the specified range.