Question

In Exercises \(35-42,\) identify the critical point and determine the local extreme values. $$y=x^{2 / 3}\left(x^{2}-4\right)$$

Short answer

The function \(y=x^{2 / 3}\left(x^{2}-4\right)\) has three critical points: \(x = 0\), which is a local minimum, and \(x = \sqrt[{2}]{{8 \over 3}}\), \(x = -\sqrt[{2}]{{8 \over 3}}\), where the exact \(y\) values depend on the evaluation of the given expression for these \(x\) values. The function does not have a local maximum.

Step-by-step solution

Differentiate the function

To find the critical points, first compute the derivative of the function using the power rule and the chain rule. The derivative of the function \(y=x^{2 / 3}\left(x^{2}-4\right)\) is \(y'= {2 \over 3}x^{-{1 \over 3}}(x^2 - 4)+x^{2 \over 3}(2x) = {2 \over 3}x^{1 \over 3}(-4) + 2x^{5 \over 3}\) which simplifies to \( y' = -{8 \over 3}x^{1 \over 3} + 2x^{5 \over 3} \)

Find the critical points

The critical points occur where the derivative is equal to zero or undefined. Set the derivative equal to zero and solve for \(x \): \(-{8 \over 3}x^{1 \over 3} + 2x^{5 \over 3} = 0\). Factor out \(x^{1 \over 3} \) to get: \( x^{1 \over 3}(-{8 \over 3}+2x^{2}) = 0\). Setting each factor equal to zero gives the critical points \(x = 0\) and \(x = \sqrt[{2}]{{8 \over 3}}\) or \(x = -\sqrt[{2}]{{8 \over 3}}\) . The derivative is also undefined at \(x = 0\), but this is already listed as a critical point.

Use the first derivative test to determine local extremities

Create a sign chart with the critical points and test the sign of the derivative in each interval. If \(x\) values plugged into the first derivative result in positive values followed by negative values, then we a have a local maxima. Conversely, negative values followed by positive values indicate a local minima. When testing, use \(x = -2, 0, 2\) to ensure they fall in the three different intervals centered around our critical points. The derivative at \(x=-2\) is positive indicating an increase, at \(x=0\) is \(0\) (a critical point), and at \(x=2\) is positive indicating an increase. Therefore, \(x=0\) is a local minimum.

Find the local extreme values

Substitute the critical points \(x = 0\) and \(x = \sqrt[{2}]{{8 \over 3}}, x = -\sqrt[{2}]{{8 \over 3}}\) into the original function to find the corresponding \(y\) values. This gives you the local extreme points; for \(x=0\), we have \(y=0\); for \(x = \sqrt[{2}]{{8 \over 3}}\), \( y = \sqrt[{2}]{{8 \over 3}}^{2 / 3}(\sqrt[{2}]{{8 \over 3}}^{2}-4)\), and for \(x = -\sqrt[{2}]{{8 \over 3}}\), \( y = -\sqrt[{2}]{{8 \over 3}}^{2 / 3}(-\sqrt[{2}]{{8 \over 3}}^{2}-4)\).