Problem 126 - Calculus Volume 1

Question

For the following exercises, find the local and/or absolute maxima for the functions over the specified domain. $$y=\sqrt{x}-\sqrt{x^{3}} \text { over }[0,4]$$

Step-by-Step Solution

Answer

The absolute maximum is \( \frac{1}{3} \) at \( x = \frac{1}{3} \).

Step 1: Calculate the Derivative

To find the local maxima, start by finding the derivative of the function. The given function is \( y = \sqrt{x} - \sqrt{x^3} \). To differentiate, rewrite it as \( y = x^{1/2} - x^{3/2} \). The derivative \( y' \) is \( \frac{1}{2}x^{-1/2} - \frac{3}{2}x^{1/2} \). Simplifying, we have \( y' = \frac{1}{2\sqrt{x}} - \frac{3}{2}x^{1/2} \).

Step 2: Find Critical Points

To find critical points, set the derivative equal to zero: \( \frac{1}{2\sqrt{x}} - \frac{3}{2}x^{1/2} = 0 \). Solve for \( x \) by multiplying through by \( 2\sqrt{x} \): \( 1 - 3x = 0 \). Solving gives \( x = \frac{1}{3} \).

Step 3: Evaluate the Function at Critical Points and Endpoints

Evaluate \( y(x) \) at the critical point \( x = \frac{1}{3} \) and endpoints \( x = 0 \) and \( x = 4 \). Calculate \( y(\frac{1}{3}) = \sqrt{\frac{1}{3}} - \sqrt{\left(\frac{1}{3}\right)^3} = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = \frac{1}{3} \). Calculate \( y(0) = \sqrt{0} - \sqrt{0^3} = 0 \). Calculate \( y(4) = \sqrt{4} - \sqrt{4^3} = 2 - 8 = -6 \).

Step 4: Determine the Maxima

Compare the values found in Step 3: \( y(0) = 0 \), \( y(\frac{1}{3}) = \frac{1}{3} \), \( y(4) = -6 \). The highest value within the domain is \( y(\frac{1}{3}) = \frac{1}{3} \). Thus, the absolute maximum over the domain \([0,4]\) is \( \frac{1}{3} \) at \( x = \frac{1}{3} \).

Key Concepts

Derivative

In calculus, a derivative represents the rate of change of a function with respect to a variable. Think of it as a way to determine how a function behaves when you make tiny changes to its input. When calculating derivatives for a function like \( y = \sqrt{x} - \sqrt{x^3} \), we start by rewriting it in a form that's easier to differentiate: \( y = x^{1/2} - x^{3/2} \). This expression is now ready for differentiation.Performing the differentiation, we apply the power rule: if \( y = x^n \), then \( y' = nx^{n-1} \). For our function:

• \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} \)
• \( \frac{d}{dx}(x^{3/2}) = \frac{3}{2}x^{1/2} \)

Thus, the derivative \( y' \) becomes \( \frac{1}{2\sqrt{x}} - \frac{3}{2}x^{1/2} \). This expression tells us how the original function changes and is especially useful when determining where it reaches peaks or valleys.

Critical Points

Critical points are specific values of \( x \) where the derivative of a function is zero or undefined. These points are significant because they can indicate local maxima or minima in the function. To find the critical points for our function, set the derivative equal to zero and solve for \( x \): \[ \frac{1}{2\sqrt{x}} - \frac{3}{2}x^{1/2} = 0 \]To simplify this equation, multiply both sides by \( 2\sqrt{x} \) to clear the fraction. Doing so yields:

• \( 1 = 3x \)

Solving for \( x \), we find that \( x = \frac{1}{3} \). This is our critical point, a potential location of a local extreme value of the function.

Evaluate the Function at Endpoints

After finding critical points, one must evaluate the function at these points and also at the endpoints of the domain to determine the absolute maxima or minima. Let's analyze our function \( y = \sqrt{x} - \sqrt{x^3} \) over the interval \([0, 4]\):Start by evaluating at the endpoints:

• At \( x = 0 \): \( y(0) = \sqrt{0} - \sqrt{0^3} = 0 \)
• At \( x = 4 \): \( y(4) = \sqrt{4} - \sqrt{4^3} = 2 - 8 = -6 \)

Next, evaluate the function at the critical point:

• At \( x = \frac{1}{3} \): \( y\left(\frac{1}{3}\right) = \sqrt{\frac{1}{3}} - \sqrt{\left(\frac{1}{3}\right)^3} = \frac{1}{\sqrt{3}} - \frac{1}{3\sqrt{3}} = \frac{1}{3} \)

Comparing these values, \( y(0) = 0 \), \( y\left(\frac{1}{3}\right) = \frac{1}{3} \), and \( y(4) = -6 \), we find the absolute maximum over the interval is \( \frac{1}{3} \) at \( x = \frac{1}{3} \). This methodical evaluation helps identify which point is the true "peak" over the given range.