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Chapter 4: Problem 61

Find the critical points, domain endpoints, and extreme values (absolute and local) for each function. $$y=x \sqrt{4-x^{2}}$$

Short Answer

Expert verified
Local max is 2 at \(x = \pm \sqrt{2}\); local min is -2 at \(x = -\sqrt{2}\); absolute min is 0 at \(x = \pm 2\).

Step by step solution

01

Determine the Domain of the Function

To find the domain of the function, consider the expression inside the square root: \(4 - x^2 \). Since we cannot have the square root of a negative number, \( 4 - x^2 \geq 0 \). This implies \(-2 \leq x \leq 2\). Hence, the domain of the function is \([-2, 2]\).
02

Find the Derivative

To find the critical points, we need to compute the derivative of \(y = x \sqrt{4 - x^2}\). Use the product rule: \( y = u \cdot v \) where \(u=x\) and \(v = \sqrt{4-x^2}\). The product rule gives \(y' = u'v + uv'\). Compute each derivative separately: \(u' = 1\) and \(v' = \frac{-x}{\sqrt{4-x^2}}\). Substitute these into the product rule to get: \( y' = 1 \cdot \sqrt{4-x^2} + x \cdot \frac{-x}{\sqrt{4-x^2}} = \frac{4-x^2 - x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}}.\)
03

Solve for Critical Points

Critical points occur where the derivative \(y'\) is zero or undefined. \(y' = 0\) when the numerator is zero: \[ 4 - 2x^2 = 0 \]Solving, we find \(2x^2 = 4\) or \(x^2 = 2\), so \(x = \pm \sqrt{2}\). Thus, the critical points are \(x = \sqrt{2}\) and \(x = -\sqrt{2}\). The derivative is undefined when \(x = \pm 2\), but these are endpoints within the domain.
04

Evaluate the Function at Critical Points and Endpoints

Calculate \(y\) at the critical points and endpoints of the domain:1. At \(x = \sqrt{2}\): \( y = \sqrt{2} \sqrt{4-(\sqrt{2})^2} = \sqrt{2} \sqrt{4-2} = \sqrt{2} \times \sqrt{2} = 2 \).2. At \(x = -\sqrt{2}\): \( y = -\sqrt{2} \sqrt{4-(\sqrt{2})^2} = -2 \).3. At \(x = 2\): \(y = 2 \sqrt{4-2^2} = 0 \).4. At \(x = -2\): \(y = -2 \sqrt{4-(-2)^2} = 0 \).
05

Identify Extreme Values

By comparing the points:- \(y = 2\) at \(x = \sqrt{2}\) is a local maximum.- \(y = -2\) at \(x = -\sqrt{2}\) is a local minimum.- \(y = 0\) at \(x = \pm 2\) are absolute minima.Thus, the extreme values include a local maximum of 2 and a local minimum of -2, with absolute minima at both endpoints.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of a Function
When working with functions, determining the domain is a crucial step. The domain tells us all the possible input values (or "x" values) that we can safely use without breaking any mathematical rules. For the function given, \(y = x \sqrt{4 - x^2}\), we need to ensure that what's inside the square root stays non-negative. In this case, the expression \(4 - x^2\) must be greater than or equal to zero.
Thus, we solve the inequality \(4 - x^2 \geq 0\) to find that the function is defined from \(-2\) to \(2\). That gives us the domain \,\([-2, 2]\). This step prevents errors like taking the square root of a negative number, which would be a misuse in real number calculations.
Derivative
To find the critical points of a function, calculating the derivative is essential. The derivative tells us the function's rate of change (or slope) at any given point. For the function \(y = x \sqrt{4-x^2}\), we use the product rule, which is perfect for dealing with functions that are multiplied together.
  • The product rule states: if you have a function \(y = u \cdot v\), then the derivative \(y'\) is \(u'v + uv'\).

Applying the product rule here entails:
  • \(u = x\) and \(v = \,\sqrt{4-x^2}\).
  • The derivatives are \(u' = 1\) and \(v' = \frac{-x}{\sqrt{4-x^2}}\).

Plug these into the formula to arrive at \(y' = \frac{4-2x^2}{\sqrt{4-x^2}}\). This expression is the derivative, indicating how the function's value changes as \(x\) changes.
Extreme Values
Extreme values of a function are points where the function reaches its highest or lowest point locally or absolutely. They are often critical points or occur at the boundaries of the domain.
For our function, critical points occur where the derivative is zero or undefined. Solving \( \frac{4 - 2x^2}{\sqrt{4-x^2}} = 0\), gives critical points at \(x = \pm \sqrt{2}\). Evaluating the function at these and at the domain's endpoints \(-2\) and \(2\), helps determine:
  • Local maximum at \(x = \sqrt{2}\) where \(y = 2\)
  • Local minimum at \(x = -\sqrt{2}\) where \(y = -2\)
  • Absolute minimum at \(x = \pm 2\) where \(y = 0\)

These extreme values characterize the function's behavior over its entire domain.
Product Rule
The product rule is a fundamental principle in calculus used to find the derivative of products of two functions. It's particularly helpful when a function is expressed as a product of simpler functions. As seen in our example, examining \(y = x \sqrt{4-x^2}\), the function can easily be divided into \(u = x\) and \(v = \sqrt{4-x^2}\).
  • The product rule formula is: \( (uv)' = u'v + uv' \).
  • This means you differentiate each part separately: \(u' = 1\), \(v' = \frac{-x}{\sqrt{4-x^2}}\).
  • Then, combine them using the rule to form the complete derivative.

This approach simplifies complex differentiation tasks and makes finding the resulting expression straightforward.

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