Problem 32 - Calculus Single Variable

Question

Find each critical point \(c\) of the given function \(f\). Then use the First Derivative Test to determine whether \(f(c)\) is a local maximum value, a local minimum value, or neither. $$ f(x)=x^{2 / 3}(x+10) $$

Step-by-Step Solution

Answer

Critical points: \( x = -10 \) is a local maximum, \( x = 0 \) is a local minimum.

Step 1: Find the First Derivative

The first step is to find the derivative of the function \( f(x) = x^{2/3}(x+10) \). We will use the product rule here. Let \( u = x^{2/3} \) and \( v = (x + 10) \). Then, \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \). Using the product rule \( (uv)' = u'v + uv' \), \[ f'(x) = \frac{2}{3}x^{-1/3}(x+10) + x^{2/3}(1). \] Simplifying this, we have: \[ f'(x) = \frac{2}{3} x^{2/3} + \frac{20}{3} x^{-1/3}. \] Combine terms: \[ f'(x) = \frac{2}{3} x^{-1/3} (x + 10). \]

Step 2: Find Critical Points

Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. From the derivative \( f'(x) = \frac{2}{3} x^{-1/3} (x + 10) \), \( f'(x) \) is undefined at \( x = 0 \) and setting \( f'(x) = 0 \), we solve \( \frac{2}{3} x^{-1/3} (x + 10) = 0 \). This is satisfied when \( x + 10 = 0 \), so \( x = -10 \). Therefore, the critical points are \( x = 0 \) and \( x = -10 \).

Step 3: Use the First Derivative Test on x = -10

To apply the First Derivative Test, check the sign of \( f'(x) \) on intervals around \( x = -10 \). Choose test points such as \( x = -11 \) and \( x = -9 \): \( f'(-11) = \frac{2}{3} (-11)^{-1/3}(-11 + 10) \) is positive, and \( f'(-9) = \frac{2}{3} (-9)^{-1/3}(-9 + 10) \) is negative. Since \( f'(x) \) changes from positive to negative, \( x = -10 \) is a local maximum.

Step 4: Use the First Derivative Test on x = 0

Check the sign of \( f'(x) \) around \( x = 0 \) by selecting points such as \( x = -1 \) and \( x = 1 \): \( f'(-1) = \frac{2}{3} (-1)^{-1/3}(-1 + 10) \) is negative, \( f'(1) = \frac{2}{3} (1)^{-1/3}(1 + 10) \) is positive. Since \( f'(x) \) changes from negative to positive, \( x = 0 \) is a local minimum.

Key Concepts

First Derivative Test

The First Derivative Test is a powerful tool used to identify local extremum points, such as local maximum and minimum values, of a function. To apply this test, you first find the critical points by setting the first derivative, \( f'(x) \), equal to zero or determining where it is undefined.

Here's how to use the test:

• Identify Critical Points: Solve \( f'(x) = 0 \) or identify where \( f'(x) \) is undefined to find your critical points.
• Choose Test Points: Select points on either side of each critical point.
• Check Sign Changes: Determine the sign of \( f'(x) \) at these test points.

These steps help us identify where the function changes from increasing to decreasing or vice versa. Hence, you'll be able to classify each critical point as a local maximum, local minimum, or neither.

Product Rule

The Product Rule is a fundamental rule in calculus used to differentiate products of two functions. When we have a function \( f(x) = u(x) v(x) \), the derivative \( f'(x) \) can be found using the Product Rule formula:

• The formula is: \[ (uv)' = u'v + uv' \]

This means that you take the derivative of the first function \( u \), multiply it by the second function \( v \), and then add the product of the first function \( u \) and the derivative of the second function \( v \).

For example, given the function \( f(x) = x^{2/3} (x + 10) \), you would let \( u = x^{2/3} \) and \( v = (x + 10) \). Differentiating, we'd find \( u' \) and \( v' \), and apply the formula to find \( f'(x) \). The Product Rule is essential when dealing with multiplication of terms that are not constants.

Local Maximum

A local maximum is a point on a function where the function reaches a highest value within a small interval around that point. This means that when you move in either direction from this point, the function's value decreases.

Using the First Derivative Test, you can identify a local maximum at a critical point as follows:

• If the derivative \( f'(x) \) transitions from positive to negative as you pass through the critical point, then the function has a local maximum at this point.

For instance, in the exercise when checking the critical point \( x = -10 \), the derivative changes from positive to negative. Hence, \( x = -10 \) is identified as a local maximum. It's like reaching the peak of a hill, where every step backward or forward leads downward.

Local Minimum

A local minimum is the opposite of a local maximum. It's a point where, within a small interval around this point, the function reaches its lowest value.

To determine a local minimum using the First Derivative Test:

• If the derivative \( f'(x) \) changes from negative to positive at a critical point, then the function attains a local minimum at that point.

In the given exercise, the critical point \( x = 0 \) shows such a behavior where the derivative transitions from negative to positive, ensuring a local minimum at that point. Imagine it as a small valley - you venture in and encounter rising terrain on both sides.