Problem 30 - 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^{1 / 3}(x+2)^{3 / 2},-2

Step-by-Step Solution

Answer

The critical point is \( x = -\frac{2}{11} \) with a local maximum at \( f(-\frac{2}{11}) \).

Step 1: Find the derivative

To find the critical points of the function \( f(x) = x^{1/3}(x+2)^{3/2} \), we first need its derivative. Use the product rule: if \( u = x^{1/3} \) and \( v = (x+2)^{3/2} \), then \( f'(x) = u'v + uv' \). Compute \( u' = \frac{1}{3}x^{-2/3} \) and \( v' = \frac{3}{2}(x+2)^{1/2} \) using the power rule.

Step 2: Apply the product rule

Using the product rule, the derivative \( f'(x) \) is: \[ f'(x) = \left(\frac{1}{3}x^{-2/3}\right)(x+2)^{3/2} + x^{1/3}\left(\frac{3}{2}(x+2)^{1/2}\right) \]. Simplify this expression to find the critical points.

Step 3: Set the derivative to zero

Set \( f'(x) = 0 \) to find potential critical points. Solve the equation \( \frac{1}{3}x^{-2/3}(x+2)^{3/2} + x^{1/3} \frac{3}{2}(x+2)^{1/2} = 0 \). Simplify and combine like terms: \( (x+2)^{1/2} \left( \frac{1}{3}(x+2) + \frac{3}{2}x \right) = 0 \).

Step 4: Solve the simplified equation

Factor out the common terms. The equation simplifies to \( (x+2)^{1/2}(\frac{x+2}{3} + \frac{3x}{2}) = 0 \). This becomes \((x+2)^{1/2}(\frac{11x + 2}{6}) = 0\). The solutions are found by setting each factor to zero: \( x+2 = 0 \) gives \( x = -2 \), but this is not within our domain \((-2, \infty)\), so it isn't considered. Now solve \( 11x + 2 = 0 \) yielding \( x = -\frac{2}{11} \).

Step 5: Apply the First Derivative Test

To apply the First Derivative Test, evaluate the sign of \( f'(x) \) around the critical point \( x = -\frac{2}{11} \). Choose \( x = -0.25 \) and \( x = 0 \) to the left and right of \( -\frac{2}{11} \). Calculate \( f'(x) \) at these points to determine if it changes sign. Compute these values to find that \( f'(x) \) changes from positive to negative, indicating a local maximum at \( x = -\frac{2}{11} \).

Key Concepts

Critical Points

Critical points are essential in understanding the behavior of a function. A critical point occurs where the derivative of a function is equal to zero or is undefined. In simpler terms, they are the points on the graph where the function's slope is flat. To find these points, you first need the derivative of the function. Once you have it, you set the derivative equal to zero and solve for the variable. It's important to remember that not all critical points will necessarily be a maximum or minimum. They are just potential locations. After identifying these points, further analysis is necessary using tests such as the First Derivative Test to determine their nature.

Derivative

The derivative of a function is a fundamental concept in calculus. It gives the rate at which the function is changing at any point. Symbolically, if you have a function \( f(x) \), the derivative is often represented as \( f'(x) \). To find this, you need to apply differentiation rules. Derivatives can be used to find critical points, understand the slope of tangents at particular points, and determine increasing or decreasing intervals of functions. In our example, differentiating the function \( f(x) = x^{1/3}(x+2)^{3/2} \) helped us pinpoint where changes in the function's behavior occurred.

Product Rule

The product rule is essential when taking derivatives of functions that are products of two other functions. If you have a function \( f(x) = u(x)v(x) \), then the derivative \( f'(x) \) is calculated as \( u'(x)v(x) + u(x)v'(x) \). This rule ensures correct differentiation by dealing with both components. In the original example, we used the product rule to differentiate \( x^{1/3} \) and \( (x+2)^{3/2} \). With this method, you compute the derivative of each component separately, then sum them up as per the rule. This approach allows for an accurate derivative calculation of a product of two functions.

Power Rule

The power rule is one of the simplest and most widely used differentiation techniques. If \( f(x) = x^n \), where \( n \) is any real number, then the derivative \( f'(x) \) is \( nx^{n-1} \). This rule is particularly advantageous when dealing with polynomial functions or any term with a power. In our function, we applied the power rule to differentiate both \( x^{1/3} \) and \( (x+2)^{3/2} \). By reducing the power by one and multiplying by the original exponent, the power rule facilitates a more straightforward and faster calculation of derivatives. This simplification aids in solving more complex calculus problems efficiently.