Question

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

Short answer

The local maximum values are at \(x = -2\) with \(y = 0\) and at \(x = 2/5\) with \(y = 312/125\). The local minimum value is at \(x = 0\) with \(y = 0\).

Step-by-step solution

Differentiate the function

Find the derivative of \(y = x^{2/3}(x+2)\) using the power rule, product rule, and chain rule if necessary. The differentiation will give \(y' = \frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3}\).

Find the critical points

Set the derivative equal to zero and solve for \(x\). The solutions to the equation \(\frac{5}{3}x^{2/3} - \frac{4}{3}x^{-1/3} = 0\) are \(x = -2\), \(x = 0\), and \(x = 2/5\). These values of \(x\) are the critical points of the function.

Determine the local extreme values

Using the first derivative test, plug in values into the derivative that are less than and greater than the critical points. From the sign of the derivative at those points, finding whether the function is increasing or decreasing around the critical points will determine if the critical points are local maxima, local minima, or neither. After testing, it is discovered that the function increases up to \(x = -2\), then decreases until \(x = 0\), then increases until \(x = 2/5\), then decreases. Therefore the local maximum occurs at \(x = -2\) and \(x =2/5\) and the local minimum occurs at \(x = 0\).