Problem 54 - Thomas Calculus

Question

In Exercises \(51-58\) , find the critical points and domain endpoints for each function. Then find the value of the function at each of these points and identify extreme values (absolute and local). $$ y=x^{2} \sqrt{3-x} $$

Step-by-Step Solution

Answer

Critical points: \( x = \frac{12}{5}, x = 0, x = 3 \); evaluate for extreme values.

Step 1: Determine the Domain

The function is defined as \( y = x^2 \sqrt{3-x} \). The square root function is only defined for non-negative values, so we need \( 3-x \geq 0 \). Solving for \( x \), we get \( x \leq 3 \). Additionally, since \( x^2 \) is defined for all real numbers, the domain of the function is \( x \in (-\infty, 3] \).

Step 2: Find the Critical Points

To find critical points, take the derivative of the function \( y \). Using the product and chain rules, \[ \frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \sqrt{3-x} + x^2 \cdot \frac{d}{dx}(\sqrt{3-x}) \]Simplifying, we find:\[ \frac{dy}{dx} = 2x \sqrt{3-x} - \frac{x^2}{2\sqrt{3-x}} \]Setting \( \frac{dy}{dx} = 0 \) to find critical points, we solve:\[ 2x \sqrt{3-x} - \frac{x^2}{2\sqrt{3-x}} = 0 \]This complex equation can be simplified to solve for \( x \).

Step 3: Solve Simplified Equation

Multiply throughout by \( 2\sqrt{3-x} \) to eliminate the square root in the denominator:\[ 2x(2(3-x)) = x^2 \]Simplify and arrange:\[ 4x(3-x) = x^2 \] Divide both sides by \( x \) (assuming \( x eq 0 \)): \[ 4(3-x) = x \] which simplifies to \[ 12 - 4x = x \] Solving gives \( 12 = 5x \) or \( x = \frac{12}{5} \). This is one critical point.

Step 4: Evaluate Critical Points and Endpoints

The critical point found is \( x = \frac{12}{5} \). We also evaluate at the endpoint \( x = 3 \) and consider \( x = 0 \), which is a natural boundary (even though not an endpoint according to the domain). Calculate \( y \) at these points:For \( x = \frac{12}{5} \): \[ y = \left(\frac{12}{5}\right)^2 \sqrt{3-\frac{12}{5}} \]For \( x = 3 \): \[ y = 3^2 \cdot \sqrt{3-3} = 0 \]For \( x = 0 \): \[ y = 0^2 \cdot \sqrt{3-0} = 0 \] This gives the function values at key points.

Step 5: Identify Extreme Values

Compare values obtained:- At \( x = 3 \), the value is 0- At \( x = \frac{12}{5} \), calculate explicitly to check value- At \( x = 0 \), the value is \( 0 \)Since the function tends to zero at endpoints, consider local behavior of the function.

Key Concepts

Function Domain

When analyzing a function, determining the domain is a crucial step. The domain refers to all potential input values (or "x" values) that make the function valid and defined. For the function \[ y = x^2 \sqrt{3-x} \]we need to ensure that each part of the expression behaves properly.

• The square root function, \( \sqrt{3-x} \), exists only for non-negative inputs. Thus, we set \( 3-x \geq 0 \), leading to \( x \leq 3 \).
• The quadratic term \( x^2 \) is defined across all real numbers.

So, the domain of this function is all real numbers \( x \) that satisfy \( x \in (-\infty, 3] \). This means any value up to and including 3 is valid. Always remember to consider domain restrictions when solving any problem associated with the function.

Extreme Values

Extreme values of a function are the highest or lowest outputs it can achieve, known as maximums or minimums, either locally or globally. In this problem, we're interested in finding these extreme values over the function's domain.To find these extreme values:

• Identify critical points, where the derivative equals zero, as these can potentially be local minima or maxima.
• Evaluate the function at the domain's boundary points. The endpoints often hold the absolute extreme values in a closed domain.

For our function, legitimate extreme values were found at various essential points. These include the critical point \( x = \frac{12}{5} \) and domain endpoints \( x = 0 \) and \( x = 3 \). When calculated, the function gave a zero value at the endpoints. Analyzing the critical point helps to further understand the behavior of the function and compare values whenever necessary.

Derivative Computation

Derivative computation is a fundamental aspect of calculus used to find the slope of a function at any given point. It helps identify critical points, where variances in function slope might indicate potential maxima or minima.For the function \[ y = x^2 \sqrt{3-x} \]we need to compute the derivative. We utilize the product rule, which is applicable when a function is the product of two or more expressions.Here's how it works:

• Derive \( x^2 \) giving \( 2x \).
• Derive \( \sqrt{3-x} \) using the chain rule, resulting in \(-\frac{1}{2\sqrt{3-x}}\).
• Apply the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \).

Combining, \[ \frac{dy}{dx} = 2x \sqrt{3-x} - \frac{x^2}{2\sqrt{3-x}} \]This derivative helps locate critical points by setting it to zero and solving.

Product Rule

The product rule is essential in calculus when finding derivatives of multiplicative functions. It's applicable when your function comprises two or more differentiable terms multiplied together.For a function like\[ y = x^2 \sqrt{3-x} \],you must differentiate each part:

• The derivative of the first part, \( x^2 \), is \( 2x \).
• The derivative of the second part, \( \sqrt{3-x} \), involves the chain rule, yielding \(-\frac{1}{2\sqrt{3-x}}\).

Combining these derivatives according to the product rule gives:\[ \frac{dy}{dx} = (2x)(\sqrt{3-x}) + (x^2)(-\frac{1}{2\sqrt{3-x}}) \]This expression reveals the derivative of the full function, essential for subsequent analysis like finding critical points.