Problem 46 - Thomas Calculus

Question

In Exercises \(45-52,\) find the derivative at each critical point and determine the local extreme values. $$ y=x^{2 / 3}\left(x^{2}-4\right) $$

Step-by-Step Solution

Answer

Critical points at \( x=1 \) and \( x=-1 \) with local extreme value of \( -3 \).

Step 1: Find the derivative of the function

First, we need to differentiate the function \( y = x^{2/3}(x^2 - 4) \). Use the product rule: \( (fg)' = f'g + fg' \), where \( f = x^{2/3} \) and \( g = x^2 - 4 \). Differentiate each component: \( f' = \frac{2}{3}x^{-1/3} \) and \( g' = 2x \). Apply the product rule to get the derivative: \[ y' = \left(\frac{2}{3}x^{-1/3}\right)(x^2 - 4) + x^{2/3}(2x). \] Simplify the expression: \[ y' = \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3}. \]

Step 2: Set the derivative equal to zero to find critical points

To find critical points, set the derivative \( y' \) equal to zero: \[ \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} = 0. \] Multiply through by \(3x^{1/3}\) to clear the fraction: \[ 2(x^2 - 4) + 6x^2 = 0. \] Simplify to get \[ 8x^2 - 8 = 0. \] Solve for \( x \): \( 8x^2 = 8 \) leading to \( x^2 = 1 \) hence \( x = \pm 1 \).

Step 3: Evaluate the derivative at critical points

Determine the sign of the derivative around the critical points to classify them. For \( x = 1 \) and \( x = -1 \), substitute back into the second derivative or test intervals around the points to classify as a local minimum or maximum.

Step 4: Determine local extreme values

Substitute \( x = 1 \) and \( x = -1 \) back into the original function \( y = x^{2/3}(x^2 - 4) \) to find the corresponding \( y \)-values: \[ y(-1) = (-1)^{2/3}((-1)^2 - 4) = 1 \times (1 - 4) = -3 \] and \[ y(1) = (1)^{2/3}((1)^2 - 4) = 1 \times (1 - 4) = -3. \] The points \( (1, -3) \) and \( (-1, -3) \) are local extrema.

Key Concepts

Derivative

In calculus, the derivative of a function measures how the function's output value changes as its input changes. Derivatives are fundamental tools for understanding how functions behave.
To find the derivative of a function like \( y = x^{2/3}(x^2 - 4) \), we often utilize derivative rules such as the product rule. The product rule helps us differentiate the product of two functions. If \( f(x) \) and \( g(x) \) are functions, then the derivative \((fg)'\) is given by:
\[ (fg)' = f'g + fg' \]In our example, \( f(x) = x^{2/3} \) and \( g(x) = x^2 - 4 \). Using the product rule:

• Find \( f' = \frac{2}{3}x^{-1/3} \)
• Find \( g' = 2x \)

Substitute into the product rule formula to get:
\[ y' = \left(\frac{2}{3}x^{-1/3}\right)(x^2 - 4) + x^{2/3}(2x) \] Simplifying gives the complete derivative. This derivative helps us explore how fast the function \( y \) is changing at any point \( x \). Calculating derivatives is crucial for finding critical points, which are intersections where changes in behavior occur.

Critical Points

Critical points of a function are where the derivative is zero or undefined. These points indicate potential local maxima, minima, or saddle points. To locate them, set the derivative \( y' \) equal to zero and solve.
In our exercise, after simplification, the equation \( y' = \frac{2}{3}x^{-1/3}(x^2 - 4) + 2x^{5/3} \) becomes:

• \( 2(x^2 - 4) + 6x^2 = 0 \)
• Simplified to \( 8x^2 - 8 = 0 \)

Solving yields \( x^2 = 1 \), giving \( x = \pm 1 \).These points, \( x = 1 \) and \( x = -1 \), are critical because they potentially signal changes in the direction of the curve from increasing to decreasing or vice versa.
It's important to evaluate the nature of these critical points by plugging values around \( x = 1 \) and \( x = -1 \) into the derivative to check the change in sign.

Local Extreme Values

Local extreme values refer to points on a function where a local maximum or minimum occurs. At these points, the function switches direction, either reaching a local high (maximum) or low (minimum).
Once critical points are identified, verify whether they are local maxima or minima. For our example:

• Calculate the original function for critical points: \( y(1) = 1\cdot(1 - 4) = -3 \)
• Similarly, \( y(-1) = 1\cdot(1 - 4) = -3 \)

Both points \( (1, -3) \) and \( (-1, -3) \) represent local extrema.
Determining if these are maxima or minima requires investigating the derivative's sign around these points. A positive-to-negative change suggests a local maximum, while a negative-to-positive change indicates a local minimum. This checking ensures accurate classification of behavior at these points.
Exploring local extreme values provides insight into a function's behavior and helps in solving optimization problems where maximizing or minimizing a result is crucial.