Problem 14 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{3}-3 x^{2}-9 x+6 $$

Step-by-Step Solution

Answer

(a) The function \(f(x) = x^3 - 3x^2 - 9x + 6\) is increasing on the intervals \((-\infty, -1)\) and \((3, \infty)\), and decreasing on the interval \((-1, 3)\). (b) There is a relative maximum at \((-1, 11)\) and a relative minimum at \((3, -21)\).

Step 1: Finding the first derivative

To find the intervals where the function is increasing or decreasing, we need to find the first derivative of the function and analyze its sign. To find the first derivative, apply the power rule: \[f'(x) = \frac{d}{dx} (x^3 - 3x^2 - 9x + 6)\] Applying the power rule, we get: \[f'(x) = 3x^2 - 6x - 9\]

Step 2: Finding critical points

Now, we need to find the critical points of the function. Critical points occur when the first derivative is equal to zero or is undefined. Since the first derivative is a polynomial, it is always defined. So, we need to find the values of \(x\) where: \[f'(x) = 3x^2 - 6x - 9 = 0\] Factoring out the common factor, we get: \[(3)(x^2 - 2x - 3) = 0\] Now, we need to find the roots of the quadratic equation \(x^2 - 2x - 3 = 0\). We can do that with factoring or the quadratic formula: \(x^2 - 2x - 3 = (x - 3)(x + 1) = 0\) So, the critical points are \(x = 3\) and \(x = -1\).

Step 3: Analyzing the intervals around the critical points

Now that we found the critical points, we need to analyze the behavior of the first derivative on the intervals around these points: I. \(x < -1\) II. \(-1 < x < 3\) III. \(x > 3\) Choose some test points from each interval and evaluate the sign of the first derivative. I. Test point: \(x = -2\) \[f'(-2) = 3(-2)^2 - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0\] So, the function is increasing in this interval. II. Test point: \(x = 1\) \[f'(1) = 3(1)^2 - 6(1) - 9 = 3 - 6 - 9 = -12 < 0\] So, the function is decreasing in this interval. III. Test point: \(x = 4\) \[f'(4) = 3(4)^2 - 6(4) - 9 = 48 - 24 - 9 = 15 > 0\] So, the function is increasing in this interval.

Step 4: Finding the relative maxima and relative minima

From the intervals analysis, we conclude that: - \(f(x)\) is increasing on \((-\infty, -1)\) and \((3, \infty)\). - \(f(x)\) is decreasing on \((-1, 3)\). Since the function changes from increasing to decreasing at \(x = -1\), there is a relative maximum at this point. We can compute the value of the function for \(x = -1\): \[f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 6 = -1 - 3 + 9 + 6 = 11\] So, there is a relative maximum at \((-1, 11)\). Similarly, since the function changes from decreasing to increasing at \(x = 3\), there is a relative minimum at this point. We can compute the value of the function for \(x = 3\): \[f(3) = (3)^3 - 3(3)^2 - 9(3) + 6 = 27 - 27 - 27 + 6 = -21\] So, there is a relative minimum at \((3, -21)\). To summarize, the given function \(f(x) = x^3 - 3x^2 - 9x + 6\) has: (a) Intervals of increasing on \((-\infty, -1)\) and \((3, \infty)\), and intervals of decreasing on \((-1, 3)\). (b) A relative maximum at \((-1, 11)\) and a relative minimum at \((3, -21)\).

Key Concepts

First Derivative

Calculus helps us understand how a function's output changes as its input changes. The first derivative of a function provides this information. It tells us the rate of change of the function at any given point.
The first derivative can be thought of as the slope of the tangent line to the function's graph.
In our example, we began by determining the first derivative of the function \(f(x) = x^3 - 3x^2 - 9x + 6\). Using the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\), we found:

• \(f'(x) = 3x^2 - 6x - 9\)

This tells us how \(f(x)\) changes with respect to \(x\). By studying its sign, we can determine where the function is increasing or decreasing.

Critical Points

Critical points occur where the first derivative is zero or undefined. These points are essential because they indicate where the function could change direction.
For polynomials, like our example function, we only need to find where the derivative equals zero because it is always defined. Setting \(3x^2 - 6x - 9 = 0\), we solve for \(x\):

• Factor to: \((x - 3)(x + 1) = 0\)
• Critical points are at \(x = 3\) and \(x = -1\)

Finding these critical points allows us to divide the number line into intervals to study the function's increasing and decreasing behavior further.

Increasing and Decreasing Intervals

An interval on a function is defined as increasing or decreasing based on the sign of the first derivative. If \(f'(x) > 0\), the function is increasing in that interval.
If \(f'(x) < 0\), it's decreasing. For the function \(f(x) = x^3 - 3x^2 - 9x + 6\), evaluating the sign of \(f'(x)\) in these intervals helps find behavior:

• Interval \((-\infty, -1)\): Increasing because test point \(x = -2\), \(f'(-2) = 15 > 0\)
• Interval \((-1, 3)\): Decreasing because test point \(x = 1\), \(f'(1) = -12 < 0\)
• Interval \((3, \infty)\): Increasing because test point \(x = 4\), \(f'(4) = 15 > 0\)

These insights are vital in sketching the graph and understanding the function's nature.

Relative Maxima and Minima

The concept of relative maxima and minima indicates points where the function reaches local high or low values. These occur at critical points where the function changes from increasing to decreasing, or vice versa. Evaluating the function at these critical points:

• At \(x = -1\), \(f(x)\) changes from increasing to decreasing (Relative Maximum): \(f(-1) = 11\).
• At \(x = 3\), \(f(x)\) changes from decreasing to increasing (Relative Minimum): \(f(3) = -21\).

This analysis provides a way to understand where the function has peaks and valleys, enabling us to better predict its behavior across different domains.