Problem 62 - Thomas Calculus in SI Units

Question

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x^{2} \sqrt{3-x}$$

Step-by-Step Solution

Answer

The absolute maximum is \( \frac{144}{25} \sqrt{\frac{3}{5}} \) at \( x = \frac{12}{5} \); the absolute minimum is 0 at \( x = 0 \) and \( x = 3 \).

Step 1: Determine the Domain

The function is given by \( y = x^2 \sqrt{3-x} \). For the square root to be real, the expression inside must be non-negative, meaning \( 3-x \geq 0 \). Solving this inequality, we find that \( x \leq 3 \). Additionally, since square roots of non-negative terms are defined, \( x \) must also be non-negative to keep \( x^2 \) nonnegative or hitting zero. Therefore, the domain of the function is \( 0 \leq x \leq 3 \).

Step 2: Find the Derivative

To find critical points, we need the derivative of \( y \). Use the product rule on \( y = x^2 \cdot (3-x)^{1/2} \). Let \( u = x^2 \) and \( v = (3-x)^{1/2} \). Then, \( u' = 2x \) and \( v' = -\frac{1}{2}(3-x)^{-1/2} \). Thus, the derivative is:\[ y' = u'v + uv' = 2x(3-x)^{1/2} + x^2 \left(-\frac{1}{2}(3-x)^{-1/2}\right) \]Simplify to:\[ y' = 2x(3-x)^{1/2} - \frac{x^2}{2}(3-x)^{-1/2} \]

Step 3: Solve for Critical Points

Set the derivative \( y' \) to zero to find critical points. The equation is:\[ 2x(3-x)^{1/2} - \frac{x^2}{2}(3-x)^{-1/2} = 0 \]Multiply through by \( 2(3-x)^{1/2} \) to clear the fraction:\[ 4x(3-x) - x^2 = 0 \]Simplifying gives:\[ 4x(3-x) = x^2 \]\[ 12x - 4x^2 = x^2 \]Rearrange to:\[ 5x^2 - 12x = 0\]Factor out \( x \):\[ x(5x - 12) = 0 \]Thus, \( x = 0 \) or \( x = \frac{12}{5} \) are the critical points.

Step 4: Evaluate the Function at Critical Points and Endpoints

Now, evaluate \( y \) at the critical points and endpoints of the domain (0 and 3):- \( x = 0 \): \( y = 0^2 \sqrt{3-0} = 0 \)- \( x = \frac{12}{5} \): \( y = \left(\frac{12}{5}\right)^2 \sqrt{3-\frac{12}{5}} = \frac{144}{25} \sqrt{\frac{3}{5}} \)- \( x = 3 \): \( y = 3^2 \sqrt{3-3} = 0 \)

Step 5: Determine Local and Absolute Extreme Values

By comparing the values of \( y \) at the critical points and domain endpoints:- At \( x = 0 \), \( y = 0 \)- At \( x = \frac{12}{5} \), \( y = \frac{144}{25} \sqrt{\frac{3}{5}} \)- At \( x = 3 \), \( y = 0 \)The absolute maximum of \( y \) is \( \frac{144}{25} \sqrt{\frac{3}{5}} \) at \( x = \frac{12}{5} \). The absolute minimum is \( 0 \) which occurs at both ends of the domain.

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial because it tells us the set of all possible input values (usually denoted as "x") for which the function is defined. For the function given by \( y = x^2 \sqrt{3-x} \), we have a square root function involved,which requires the expression inside the square root, \( 3-x \), to be non-negative. This means:

• \( 3-x \geq 0 \)
• \( x \leq 3 \)

Furthermore, since there is \( x^2 \), \( x \) must be non-negative to not cause the square term to take on negative values, establishing:

• \( x \geq 0 \)

Combining both conditions, the domain of the function is:

• \( 0 \leq x \leq 3 \)

This domain ensures that the function is only evaluated at points where it remains real and valid.

Extreme Values

Extreme values of a function are critical as they represent the highest or lowest values the function can attain. These points are split into local extremes, which occur within a specific subset of the domain, and absolute extremes,which are the highest and lowest values within the entire domain of the function.
To find the extreme values, we first determine the critical points where the derivativeis zero or undefined.
In our function:

• The endpoints of the domain are \( x = 0 \) and \( x = 3 \).
• The critical point within the domain, from solving the derivative, is \( x = \frac{12}{5} \).

Evaluating the function at these points helps us locate the extreme values:

• \( y(0) = 0 \): a minimum at one endpoint.
• \( y \left(\frac{12}{5}\right) = \frac{144}{25} \sqrt{\frac{3}{5}} \): the absolute maximum of the function.
• \( y(3) = 0 \): a minimum at the other endpoint.

This means the function varies between 0 (minimum) and a maximum value at \( x = \frac{12}{5} \).

Derivative and Differentiation

Differentiation helps us understand how a function changes with respect to its input. In other words, it gives us the rate of change of the function, and is key to finding critical points for determining extreme values.
The derivative of a function, denoted as \( y' \), is calculated usingrules like the power rule and the product rule, as applicable.
For \( y = x^2 \sqrt{3-x} \), we can apply the product rulebecause it involves the product of two functions:

• Let \( u = x^2 \) and \( v = (3-x)^{1/2} \)

The derivative of the product \( uv \) is:

• \( y' = u'v + uv' \)

For these components:

• \( u' = 2x \) and \( v' = -\frac{1}{2}(3-x)^{-1/2} \)

Thus,\[y' = 2x(3-x)^{1/2} - \frac{x^2}{2}(3-x)^{-1/2}\]This simplified derivative helps us pinpoint critical points to analyze the function's behavior.

Product Rule

The product rule is a fundamental tool used in differentiation when dealing with products of two functions. If a function \( y \) can be expressed as a product of two differentiable functions, say \( u \) and \( v \), then the product rule states:

• \( y' = u'v + uv' \)

This rule is especially useful in the differentiation of our function, \( y = x^2 \sqrt{3-x} \).
Here, \( u = x^2 \) and \( v = (3-x)^{1/2} \), as defined:

• Differentiate each: \( u' = 2x \) and \( v' = -\frac{1}{2}(3-x)^{-1/2} \).
• Apply the product rule: \( y' = 2x(3-x)^{1/2} + x^2 \left(-\frac{1}{2}(3-x)^{-1/2}\right) \).

By simplifying this expression, we obtain the complete derivative,critical for finding where the function's slope is zero,an essential step for analyzing extreme and critical points.