Problem 226 - Calculus Volume 1

Question

For the following exercises, determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). \(f(x)=x^{4}-6 x^{3}\)

Step-by-Step Solution

Answer

\(f\) is decreasing on \((-\infty, \frac{9}{2})\) and increasing on \((\frac{9}{2}, \infty)\); \(f\) has a local minimum at \(x = \frac{9}{2}\).

Step 1: Find the first derivative

To determine where the function is increasing or decreasing, we first need to find the first derivative of the function. Given the function is \(f(x) = x^4 - 6x^3\), the first derivative is found by differentiating: \(f'(x) = 4x^3 - 18x^2\).

Step 2: Find critical points

Critical points occur where the first derivative is zero or undefined. Set \(f'(x) = 0\): \(4x^3 - 18x^2 = 0\). Factor the equation: \(x^2(4x - 18) = 0\). This gives \(x^2 = 0\) or \(4x - 18 = 0\). Thus, critical points are \(x = 0\) and \(x = \frac{9}{2}\).

Step 3: Determine intervals of increase and decrease

Use the critical points to test the sign of \(f'(x)\) in intervals: \((-\infty, 0)\), \((0, \frac{9}{2})\), and \((\frac{9}{2}, \infty)\). Choose test points such as \(-1\), \(1\), and \(5\) and evaluate \(f'(x)\): - For \(x = -1\), \(f'(-1) = 4(-1)^3 - 18(-1)^2 = -4 - 18 = -22\), so \(f\) is decreasing. - For \(x = 1\), \(f'(1) = 4(1)^3 - 18(1)^2 = 4 - 18 = -14\), so \(f\) is decreasing.- For \(x = 5\), \(f'(5) = 4(5)^3 - 18(5)^2 = 500 - 450 = 50\), so \(f\) is increasing.Thus, \(f\) is decreasing on \((-\infty, \frac{9}{2})\) and increasing on \((\frac{9}{2}, \infty)\).

Step 4: Find local minima and maxima

Since the function changes from decreasing to increasing at \(x = \frac{9}{2}\), there is a local minimum at \(x = \frac{9}{2}\). Evaluate \(f(x)\) at this point: \(f\left(\frac{9}{2}\right) = \left(\frac{9}{2}\right)^4 - 6\left(\frac{9}{2}\right)^3 = \frac{6561}{16} - \frac{4374}{8}\). Simplifying, this results in a local minimum value of \(-\frac{729}{8}.\)

Key Concepts

Derivatives

Derivatives are fundamental in understanding the behavior of functions. In our exercise, we have the function \(f(x) = x^4 - 6x^3\). To find out how this function behaves, we start by finding the first derivative. This derivative shows how the function's value changes as \(x\) changes. It's noted as \(f'(x)\). Differentiation is the process of finding a derivative. We differentiate each term in the function separately:

• The derivative of \(x^4\) is \(4x^3\),
• The derivative of \(-6x^3\) is \(-18x^2\).

This gives us \(f'(x) = 4x^3 - 18x^2\). We'll use this derivative in the following sections to analyze the function further.

Critical Points

Critical points of a function are where its derivative is zero or undefined. These points help us identify potential local maxima and minima.For the function \(f(x) = x^4 - 6x^3\), we find the critical points by setting the derivative \(f'(x) = 4x^3 - 18x^2\) to zero. This involves solving the equation: \[4x^3 - 18x^2 = 0.\] Factoring out \(x^2\), we get \[x^2(4x - 18) = 0.\] This results in \(x = 0\) or \(4x - 18 = 0\), which gives us the critical points \(x = 0\) and \(x = \frac{9}{2}\). These points are crucial for exploring where the function's behavior changes.

Increasing and Decreasing Intervals

Once we have the critical points, we analyze intervals around these points to determine where the function is increasing or decreasing. This is done by choosing test points in each interval and calculating the sign of \(f'(x)\).Consider the intervals

• \((-\infty, 0)\),
• \((0, \frac{9}{2})\),
• \((\frac{9}{2}, +\infty)\).

For example, testing points \(-1\), \(1\), and \(5\) gives:

• \(f'(-1) = -22\), so the function decreases on \((-\infty, 0)\).
• \(f'(1) = -14\), so the function decreases on \((0, \frac{9}{2})\).
• \(f'(5) = 50\), so the function increases on \((\frac{9}{2}, +\infty)\).

Using these intervals, we find that \(f\) decreases in the intervals \((-\infty, \frac{9}{2})\) and increases in \((\frac{9}{2}, +\infty)\), indicating a local minimum at \(x = \frac{9}{2}\).

Concavity and Inflection Points

Concavity describes how curved a function's graph is, telling us whether the function is bending upwards (concave up) or downwards (concave down). Inflection points are where the concavity changes.To analyze concavity, we find the second derivative. So, differentiating \(f'(x) = 4x^3 - 18x^2\) gives us the second derivative: \[f''(x) = 12x^2 - 36x.\] Setting \(f''(x) = 0\) helps us find potential inflection points:\[12x^2 - 36x = 0.\] Factoring out \(12x\), we have \[12x(x - 3) = 0.\] Thus, \(x = 0\) and \(x = 3\) are candidates for inflection points.Analyzing intervals around these points,

• Choosing test values gives insight into the sign changes of \(f''(x)\),
• Where changes from positive to negative or vice versa indicate inflection points.

For instance, if \(f''(x)\) changes from positive to negative at \(x = 3\), the graph changes from concave up to concave down.