Problem 28 - Calculus A New Horizon

Question

Use any method to find the relative extrema of the function \(f\). $$f(x)=2 x+x^{2 / 3}$$

Step-by-Step Solution

Answer

The critical point \(x = -1/27\) is a local maximum.

Step 1: Find the derivative of the function

To find the relative extrema, we first need the derivative of the function, as critical points occur where the derivative is zero or undefined. The function is \(f(x) = 2x + x^{2/3}\). Applying the power rule, we find: \(f'(x) = 2 + \frac{2}{3}x^{-1/3}\).

Step 2: Find critical points

Set the derivative equal to zero to find critical points, \(2 + \frac{2}{3}x^{-1/3} = 0\). Solve for \(x\): \[2 = -\frac{2}{3}x^{-1/3}\] Multiply both sides by \(-3\): \[ -6 = 2x^{-1/3} \] \[ x^{-1/3} = -3 \] Raise both sides to the power of \(-3\): \[ x = (-3)^{-3} = -1/27 \].

Step 3: Evaluate derivative at the critical point

To determine the nature of the critical point found, compute the second derivative. First, find the derivative of \(f'(x) = 2 + \frac{2}{3}x^{-1/3}\), which gives \( f''(x) = \frac{-2}{9}x^{-4/3} \).

Step 4: Analyze the second derivative

Evaluate the second derivative \(f''(x)\) at \(x = -1/27\): \[f''(-1/27) = \frac{-2}{9}(-1/27)^{-4/3} \]The exponent \(-4/3\) will make negative \(-1/27\) result in a real negative number, so \(f''(-1/27) < 0\), indicating that \(x = -1/27\) is a local maximum.

Key Concepts

Understanding Derivatives

Derivatives play a crucial role in calculus, providing us insights into the functionality of equations. When you compute a derivative, you're essentially finding the slope of the tangent line at any point on a curve.

• The derivative gives us the rate of change of a function's value with respect to changes in the independent variable, usually denoted as 'x.'
• Finding the derivative involves various rules, with the power rule being one of the most commonly used: if you have a function like \(x^n\), its derivative is \(nx^{n-1}\).

In the exercise, the derivative of \(f(x) = 2x + x^{2/3}\) is found using this rule, resulting in \(f'(x) = 2 + \frac{2}{3}x^{-1/3}\). By calculating this, you're preparing to identify critical points that will uncover where the function changes its slope dramatically, indicating possible relative extrema.

Identifying Relative Extrema

Relative extrema, including both relative maxima and minima, are points where the function reaches a peak or a trough within an interval. These points are significant in understanding the behavior of the function.

• A maximum point is where the function changes from increasing to decreasing, forming a 'hilltop' on the graph.
• A minimum occurs where the function changes from decreasing to increasing, creating a 'valley.'

To find these points, we first look for critical points by setting the first derivative to zero (or finding where it's undefined). These critical points are possible candidates for relative extrema. In our exercise, we set \(2 + \frac{2}{3}x^{-1/3} = 0\) to find the critical point \(x = -1/27\). Once identified, it's important to verify their nature through further testing, often using the second derivative.

Applying the Second Derivative Test

The Second Derivative Test is a handy tool to confirm the nature of critical points found using the first derivative. It utilizes the second derivative of the function, \(f''(x)\), to analyze the concavity at critical points.

• If \(f''(c) > 0\), the function is concave up at \(x = c\), and \(c\) is a local minimum.
• If \(f''(c) < 0\), the function is concave down at \(x = c\), and \(c\) is a local maximum.
• If \(f''(c) = 0\), the test is inconclusive.

In our case, the second derivative, \(f''(x) = \frac{-2}{9}x^{-4/3}\), was calculated. When evaluated at the critical point \(x = -1/27\), we found \(f''(-1/27) < 0\), indicating that \(x = -1/27\) is indeed a local maximum. This test is an efficient method to sort through critical points and confirm their respective roles as relative extrema.