Problem 22 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x\left(\frac{x}{2}-5\right)^{4} \end{equation}

Step-by-Step Solution

Answer

Find critical points using first derivative, use the second derivative for extrema and inflection points, and graph the function.

Step 1: Expand and Simplify the Function

First, rewrite the function by expanding the term. Given \( y = x \left( \frac{x}{2} - 5 \right)^4 \), let's expand the expression \( \left( \frac{x}{2} - 5 \right)^4 \). Use the binomial expansion or directly expand to simplify the function for differentiation.

Step 2: Take the First Derivative

Find the first derivative \( y' \) to identify critical points. Use the product rule for derivatives. Let \( u = x \) and \( v = \left( \frac{x}{2} - 5 \right)^4 \), then \( y' = u'v + uv' \). Compute \( u' \) and \( v' \) separately.

Step 3: Solve for Critical Points

Set the first derivative \( y' \) to zero: \( u'v + uv' = 0 \). Solve this equation for \( x \) to determine the critical points where local extremum might occur.

Step 4: Use the Second Derivative Test

Find the second derivative \( y'' \) to apply the second derivative test to the critical points identified in Step 3. Determine whether the points are local minima or maxima.

Step 5: Identify Inflection Points

Take the second derivative \( y'' \) and set it to zero to find possible inflection points. Check the sign change around these values to confirm if they are indeed inflection points.

Step 6: Graph the Function

Using the findings from the derivative tests, graph the function \( y = x \left( \frac{x}{2} - 5 \right)^4 \) to visualize the local and absolute extrema and any points of inflection. Include axis intersection points for reference.

Key Concepts

Critical Points

Critical points are where function changes can occur, such as local maxima, minima, or saddles. To find these, we need to first compute the function's derivative. For a function like \( f(x) = x \left( \frac{x}{2} - 5 \right)^4 \), the critical points are the \( x \) values where the first derivative is zero or undefined. After finding the derivative, set it equal to zero and solve for \( x \). These solutions are your potential critical points. Evaluate the function value at these \( x \) points to determine if they are actual extrema. Remember, not all critical points are extrema; some may simply indicate a point of inflection where the curvature of the graph changes.

Binomial Expansion

The binomial expansion is a way to expand expressions that are raised to a power, typically written in the form \((a+b)^n\). In cases where you have a compound expression like \( \left( \frac{x}{2} - 5 \right)^4 \), binomial expansion helps transform it into a more manageable polynomial for easier differentiation. The general formula for binomial expansion is:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

You'd replace \(a\) and \(b\) with \(\frac{x}{2}\) and \(-5\) respectively, and then calculate each term. This provides a polynomial form of the original expression, allowing for straightforward application of derivative rules.

First Derivative

The first derivative of a function tells us the rate at which the function value changes with respect to changes in \( x \). For our function, using the product rule makes differentiation straightforward. The product rule for two components \( u \) and \( v \) is \( (uv)' = u'v + uv' \). Applying this to our function:

• Choose \( u = x \) and \( v = \left( \frac{x}{2} - 5 \right)^4 \)
• Compute \( u' \) as 1 and \( v' \) using the chain rule \( 4 \left( \frac{x}{2} - 5 \right)^3 \times \frac{1}{2} \)
• Combine to get \( y' = 1 \cdot v + x \cdot v' \)

Solve the resulting equation to find critical points. The first derivative changes sign at these points, indicating possible extrema.

Second Derivative Test

The second derivative test helps to determine the nature of critical points identified by the first derivative. Once you've found potential critical points, you can use the second derivative to decide if these points are maxima, minima, or neither. Compute the second derivative by differentiating the first derivative. Plug the critical points into this new derivative:

• If \( y''(x) > 0 \), the function is concave up, indicating a local minimum.
• If \( y''(x) < 0 \), the function is concave down, indicating a local maximum.
• If \( y''(x) = 0 \), the test is inconclusive; further testing is needed.

This test offers more precision in deciphering the nature of each critical point.