Problem 32 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x^{2 / 3}(x-4)^{1 / 3} $$

Step-by-Step Solution

Answer

The function is concave up on \( (\frac{8}{3}, 4) \) and concave down on \( (0, \frac{8}{3}) \). The point of inflection is at \( x = \frac{8}{3} \). The point \( x = \frac{8}{3} \) is a local minimum.

Step 1: Find the First Derivative

To find critical points and analyze the concavity, first compute the first derivative of the function. Apply the product rule to \( f(x) = x^{2/3}(x-4)^{1/3} \). This gives: \[ f'(x) = \frac{d}{dx}[x^{2/3}](x-4)^{1/3} + x^{2/3}\frac{d}{dx}[(x-4)^{1/3}]. \] Calculating each derivative using the power rule: \[ \frac{d}{dx}[x^{2/3}] = \frac{2}{3}x^{-1/3}, \] \[ \frac{d}{dx}[(x-4)^{1/3}] = \frac{1}{3}(x-4)^{-2/3}. \] Now substitute back: \[ f'(x) = \frac{2}{3}x^{-1/3}(x-4)^{1/3} + \frac{1}{3}x^{2/3}(x-4)^{-2/3}. \] Simplify the expression. Identify and equate \( f'(x) \) to zero to find critical points.

Step 2: Simplify and Find Critical Points

Simplify the first derivative by factoring common terms: \[ f'(x) = \frac{1}{3}x^{-1/3}(x-4)^{-2/3} [2(x-4) + x]. \] Simplify the expression inside brackets: \[ 2(x-4) + x = 2x - 8 + x = 3x - 8. \] So, \[ f'(x) = \frac{1}{3}x^{-1/3}(x-4)^{-2/3}(3x - 8). \] Set \( f'(x) = 0 \) to find critical points: \( 3x - 8 = 0 \), solve for \( x \): \[ x = \frac{8}{3}. \] Check where \( x^{2/3}(x-4)^{1/3} \) might be undefined, noting \( x=0 \) and \( x=4 \) are points to consider separately.

Step 3: Find the Second Derivative

For concavity and points of inflection, compute the second derivative. From \[ f'(x) = \frac{1}{3}x^{-1/3}(x-4)^{-2/3}(3x - 8), \] compute \[ f''(x). \] Differentiate term by term. This requires the use of the product rule and chain rule, breaking into components and handling each separately.

Step 4: Determine Concavity and Points of Inflection

Using the second derivative \( f''(x) \), determine where it changes sign: find where \( f''(x) = 0 \) or is undefined, checking intervals around these values to establish concavity. Points where sign changes indicate inflection points. Calculate \( f''(x) \) at test points to identify intervals of concavity.

Step 5: Apply Second Derivative Test for Local Extrema

Use the second derivative test on the critical point \( x = \frac{8}{3} \). Check the sign of \( f''(x) \) at \( x = \frac{8}{3} \): - If \( f''(x) > 0 \), \( x \) is a local minimum.- If \( f''(x) < 0 \), \( x \) is a local maximum.

Key Concepts

Critical Points

In calculus, understanding critical points is crucial when analyzing functions. Critical points are those points on the graph of a function where the derivative is zero or undefined. These points indicate where the graph might have a local minimum, local maximum, or a point of inflection. When dealing with polynomial functions, you compute the first derivative of the function and solve the equation \( f'(x) = 0 \) or find where \( f'(x) \) is undefined to identify critical points.
In the exercise provided, the critical point is found at \( x = \frac{8}{3} \), where the first derivative has a solution. Additionally, points like \( x = 0 \) and \( x = 4 \) are considered due to potential undefined behavior in the function. This is crucial for understanding the behavior of the function around these points.

Second Derivative Test

The second derivative test is a method used to determine whether a critical point is a local minimum, local maximum, or a point of inflection. Simply put, you calculate the second derivative of the function, \( f''(x) \), and evaluate it at each critical point.

• If \( f''(x) > 0 \) at a critical point, the function is concave up there, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis might be needed to determine the point's nature.

In the context of the problem, you apply this test to \( x = \frac{8}{3} \) to verify whether it's a local minimum or maximum. Depending on the sign of the second derivative at this point, you can classify the nature of the critical point confidently.

Points of Inflection

Points of inflection are points on the graph of a function where the concavity changes. That means the curve changes from being concave up to concave down, or vice versa. To find points of inflection, you need to examine the second derivative \( f''(x) \).

• A point of inflection exists where \( f''(x) = 0 \) and the sign of \( f''(x) \) changes as you pass through this point.
• Additionally, a point could be a point of inflection if \( f''(x) \) is undefined but changes sign around it.

In the given problem, identifying where the second derivative is zero or changes sign is essential for pinning down the points of inflection. This tells you where the curvature of the graph shifts direction, giving a complete picture of the function's behavior.

Polynomial Functions

Polynomial functions are algebraic expressions including terms up to a certain power. These functions are smooth and continuous, characterized by their degree which determines the number of roots or solutions, and critical points.

• The degree of a polynomial directly affects the shape and complexity of its graph.
• They can have multiple critical points, local minima, maxima, and inflection points depending on their degree and coefficients.
• The first and second derivatives of polynomial functions help in finding these points and understanding changes in the graph's direction.

In this exercise, the function \( f(x) = x^{2/3}(x-4)^{1/3} \) is not a classic polynomial due to its fractional powers but exhibits similar behavior allowing such analysis. Recognizing these functions' derivative properties is crucial for identifying key features such as critical points and inflection points.