Problem 20 - Calculus Early Transcendentals

Question

Find: (a) the intervals on which f is increasing, (b) the intervals on which f is decreasing, (c) the open intervals on which f is concave up, (d) the open intervals on which f is concave down, and (e) the x-coordinates of all inflection points. \(f(x)=x^{4}-5 x^{3}+9 x^{2}\)

Step-by-Step Solution

Answer

(a) Increasing on \((\frac{3}{2}, \infty)\); (b) Decreasing on \((-\infty, 0)\) and \((0, \frac{3}{2})\); (c) Concave up on \((\frac{3}{2}, \infty)\); (d) Concave down on \((-\infty, 1)\) and \((1, \frac{3}{2})\); (e) Inflection points at \(x=1\) and \(x=\frac{3}{2}\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first need to find the derivative of the function. Calculate \(f'(x)\) by differentiating \(f(x) = x^4 - 5x^3 + 9x^2\). This gives \(f'(x) = 4x^3 - 15x^2 + 18x\).

Step 2: Set the First Derivative to Zero

Find the critical points where the function could change from increasing to decreasing or vice versa by setting \(f'(x) = 0\). Thus, solve \(4x^3 - 15x^2 + 18x = 0\). Factor the equation: \(x(2x - 3)(2x - 3) = 0\). The solutions are \(x = 0\) and \(x = \frac{3}{2}\).

Step 3: Determine Intervals of Increase and Decrease

Test intervals around the critical points to determine where \(f(x)\) is increasing or decreasing. Use test points in the intervals \((-\infty, 0), (0, \frac{3}{2}), (\frac{3}{2}, \infty)\).- In \((-\infty, 0)\), choose \(x = -1\). Evaluate \(f'(-1)\). If the value is positive, the function is increasing.- In \((0, \frac{3}{2})\), choose \(x = 1\). Evaluate \(f'(1)\).- In \((\frac{3}{2}, \infty)\), choose \(x = 2\). Evaluate \(f'(2)\).

Step 4: Find the Second Derivative

To find concavity, calculate the second derivative: \(f''(x) = 12x^2 - 30x + 18\).

Step 5: Set the Second Derivative to Zero

Find where the concavity changes by solving \(f''(x) = 0\), which means solving \(12x^2 - 30x + 18 = 0\). Divide by 6: \(2x^2 - 5x + 3 = 0\), and factor it to \((x-1)(2x-3) = 0\). The solutions are \(x = 1\) and \(x = \frac{3}{2}\).

Step 6: Determine Intervals of Concavity

Determine the sign of \(f''(x)\) in the intervals \((-\infty, 1), (1, \frac{3}{2}), (\frac{3}{2}, \infty)\). - In \((-\infty, 1)\), choose \(x = 0\).- In \((1, \frac{3}{2})\), choose \(x = 1.25\).- In \((\frac{3}{2}, \infty)\), choose \(x = 2\).

Step 7: Identify Inflection Points

Inflection points occur where the concavity changes. From the solution to \(f''(x) = 0\), check \(x = 1\) and \(x = \frac{3}{2}\). Confirm if there is a change in concavity at these x-values.

Key Concepts

Critical Points

Critical points are x-values where the function's derivative equals zero, or where it is undefined. These points indicate potential changes in the function's behavior from increasing to decreasing, or vice versa. For the given function, the critical points are discovered by setting the first derivative equal to zero. This involves solving the equation for critical points.
To find critical points, first, calculate the first derivative of the function. In the given problem, the first derivative is:

• \(f'(x) = 4x^3 - 15x^2 + 18x\)

Setting the derivative equal to zero gives the equation:

• \(4x^3 - 15x^2 + 18x = 0\)

By factoring, we find:

• \(x(2x - 3)(2x - 3) = 0\)

Solving this provides the critical points at \(x = 0\) and \(x = \frac{3}{2}\).
These critical points are useful in determining where the function increases or decreases on a graph.

First Derivative

The first derivative of a function is used to find where the function is increasing or decreasing. By examining the sign of the first derivative in various intervals around critical points, you can determine the behavior of the function.
To determine if the function is increasing or decreasing, inspect the first derivative at test points in the intervals formed: \((-\infty, 0)\), \((0, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).

• Choose a test point in \((-\infty, 0)\), such as \(x = -1\). The derivative at this point shows whether the function is increasing.
• In \((0, \frac{3}{2})\), use \(x = 1\) to verify the function's behavior through the derivative.
• For the interval \((\frac{3}{2}, \infty)\), check \(x = 2\).

In short, if the derivative is positive in an interval, the function is increasing there. Conversely, if the derivative is negative, the function is decreasing.

Second Derivative

The second derivative provides information on the concavity of a function, which can help identify inflection points. Concavity tells us if the graph curves upwards or downwards in certain intervals.
For the given function, the second derivative is calculated as:

• \(f''(x) = 12x^2 - 30x + 18\)

Set the second derivative to zero to find possible inflection points:

• \(12x^2 - 30x + 18 = 0\)
• Simplified, it becomes \((x - 1)(2x - 3) = 0\)

This gives the solutions \(x = 1\) and \(x = \frac{3}{2}\). These points might indicate changes in the graph's concavity.

Intervals of Concavity

To ascertain the intervals where the function is concave up or down, assess the second derivative at points within the derived intervals \((-\infty, 1)\), \((1, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).

• In the interval \((-\infty, 1)\), try \(x = 0\). Determine whether \(f''(0)\) is positive or negative to check if the function is concave up (positive) or concave down (negative).
• Do the same for \((1, \frac{3}{2})\) with \(x = 1.25\).
• Lastly, for \((\frac{3}{2}, \infty)\), use \(x = 2\).

If the second derivative is positive, the function is concave up; if negative, it's concave down. Changes in concavity at \(x = 1\) and \(x = \frac{3}{2}\) indicate inflection points where the curve's direction changes.