Problem 33 - Thomas Calculus

Question

Use the steps of the graphing procedure on page 272 to graph the equations in Exercises \(9-40 .\) Include the coordinates of any local extreme points and inflection points. $$ y=x \sqrt{8-x^{2}} $$

Step-by-Step Solution

Answer

Critical points: \((2,4)\) local max, \((-2,-4)\) local min; inflection point at \((0,0)\).

Step 1: Determine the Domain

To find the domain of the function \( y = x \sqrt{8-x^2} \), we look for values of \( x \) such that \( 8-x^2 \geq 0 \). Solving \( x^2 \leq 8 \), we get \( -\sqrt{8} \leq x \leq \sqrt{8} \). Hence, the domain is \( [-\sqrt{8}, \sqrt{8}] \).

Step 2: Find the Derivative

To find critical points, we first calculate the derivative of \( y = x \sqrt{8-x^2} \). Let \( u = x \) and \( v = \sqrt{8-x^2} \). Using the product rule \( (uv)' = u'v + uv' \), we have \( u' = 1 \) and \( v' = \frac{-x}{\sqrt{8-x^2}} \). Thus, \( y' = \sqrt{8-x^2} + x \cdot \frac{-x}{\sqrt{8-x^2}} = \frac{8 - 2x^2}{\sqrt{8-x^2}} \).

Step 3: Find Critical Points

Set the derivative \( y' = \frac{8 - 2x^2}{\sqrt{8-x^2}} = 0 \). Solving \( 8 - 2x^2 = 0 \) yields \( x = \pm 2 \). So, the critical points are \( x = 2 \) and \( x = -2 \).

Step 4: Calculate Second Derivative

To classify the critical points, find the second derivative using \( y'' = \left( \frac{8 - 2x^2}{\sqrt{8-x^2}} \right)' \). This requires applying the quotient rule and simplifying to determine if it changes sign at critical and other points.

Step 5: Classify and Evaluate Critical Points

Evaluating \( y(x) \) at critical points for extreme values: \( y(2) = 2\sqrt{4} = 4 \) and \( y(-2) = -4 \); hence, \((2, 4)\) is a local maximum and \((-2, -4)\) is a local minimum. Second derivative test confirms these classifications.

Step 6: Identify Inflection Points

Set \( y'' = 0 \) and solve to find \( x eq -\sqrt{8}, 0, \sqrt{8} \) where signs of \( y' \) change. This yields \( x = 0 \) indicating it is an inflection point with \( y(0) = 0 \).

Step 7: Graph the Function

Using the domain and critical points, plot \( y = x \sqrt{8-x^2} \) with endpoints \((\pm\sqrt{8},0)\), local extrema at \((2, 4), (-2, -4)\), and an inflection point at \((0,0)\). This completes the graph with symmetry about the origin.

Key Concepts

Derivative

When graphing a function, understanding the derivative is crucial. The derivative of a function gives us the slope of the tangent line to the curve at any given point. This slope indicates how the function is changing at that point.

• The first derivative, often denoted as \( y' \) or \( f'(x) \), helps identify critical points.
• For the function \( y = x \sqrt{8-x^2} \), we apply the product rule to find the derivative: \( y' = \frac{8 - 2x^2}{\sqrt{8-x^2}} \).

Understanding the derivative is essential for finding where the function has local maxima or minima. Knowing how to differentiate effectively is key to analyzing the behavior of functions.

Critical Points

Critical points of a function are where the derivative is zero or undefined. These points help us pinpoint where the function may reach local maximum or minimum values.

• For \( y = x \sqrt{8-x^2} \), the critical points occur where \( y' = 0 \). Solving \( \frac{8 - 2x^2}{\sqrt{8-x^2}} = 0 \) yields \( x = \pm 2 \).
• At these points, the function evaluates to \((2, 4)\) and \((-2, -4)\), indicating a local maximum and minimum respectively.

Critical points are valuable for sketching the graph of the function, as they indicate where the graph changes direction.

Inflection Points

Inflection points occur where the second derivative changes sign. This indicates where the curve changes its concavity, which means transitioning from concave upwards to concave downwards, or vice versa.

• To find these points, solve \( y'' = 0 \) for your function.
• For the function \( y = x \sqrt{8-x^2} \), solving \( y'' = 0 \) gives us the inflection point at \( x = 0 \).
• This point \((0, 0)\) shows where the graph of the function switches concavity, an important feature when understanding the curvature of the graph.

Recognizing inflection points gives deeper insight into the shape and behavior of the graph.

Domain of a Function

The domain of a function consists of all the possible input values (\( x \)-values) that will produce a valid output. It's crucial to determine this before embarking on graphing or solving a function.

• For \( y = x \sqrt{8-x^2} \), the domain is defined by ensuring the expression under the square root is non-negative: \( 8-x^2 \geq 0 \).
• This results in the domain being \([-\sqrt{8}, \sqrt{8}]\).

Knowing the domain is fundamental as it tells us the range over which the function can be graphed and correctly analyzed. It's the first step in understanding the limitations and potential of any function.