Problem 23 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=(x-1)^{2}(x+3) $$

Step-by-Step Solution

Answer

The critical numbers of \(f(x)\) are \(x = -5/3\) and \(x = 1\). The function is increasing on \((-\infty, -5/3)\) and \((1, \infty)\) and decreasing on \((-5/3, 1)\). The relative maximum is at \(x = -5/3\) and the relative minimum is at \(x = 1\).

Step 1: Find the derivative of the function

The first step is to find the first derivative of the function \(f(x) = (x-1)^{2}(x+3)\). Using the power rule and chain rule for differentiation, \[ f'(x) = 2(x-1)(x+3) + (x-1)^2 \]. This simplifies to \(f'(x) = 3x^{2} - 2x - 5\).

Step 2: Find the critical numbers

The critical numbers are found by setting the derivative equal to zero and solving for \(x\), \[0 = 3x^{2} - 2x - 5\]. Solving this quadratic gives \(x = 1\) and \(x = -5/3\) meaning these are the critical numbers.

Step 3: Determine where the function is increasing or decreasing

Using a number line and the critical numbers from Step 2, test intervals to see where the derivative is positive or negative. If \(x < -5/3\), \(f'(x) > 0\), meaning the function is increasing on the interval \((-\infty, -5/3)\). Between \( -5/3 < x < 1\), \(f'(x) < 0\), indicating the function is decreasing on the interval \((-5/3, 1)\). Finally, for \(x > 1\), \(f'(x) > 0\), so the function is increasing on the interval \((1, \infty)\).

Step 4: Apply the First Derivative Test

From the observations in the previous step, at \(x = -5/3\), the function changes from increasing to decreasing, so \(-5/3\) is a relative maximum. At \(x = 1\), the function changes from decreasing to increasing, so \(1\) is a relative minimum.

Step 5: Graph the function

Graph the function using a graphing utility and confirm that there is a relative maxima at \(x = -5/3\) and a relative minima at \(x = 1\). The graph will have a peak at \(x = -5/3\) and a valley at \(x = 1\), confirming the results found above.

Key Concepts

Critical Numbers

Understanding critical numbers is essential when analyzing the behavior of a function. These numbers are found by taking the derivative of a function and setting it equal to zero. The solutions to this equation are the function's critical numbers, which could indicate potential local maxima, minima, or points of inflection.

For the given function, the analysis began with finding the derivative, which was simplified to \(f'(x) = 3x^{2} - 2x - 5\). Setting this equal to zero gave the critical numbers \(x = 1\) and \(x = -\frac{5}{3}\). It's important to note that in addition to these, any points where the derivative does not exist, or the function is undefined, can also be counted as critical numbers.

Increasing and Decreasing Intervals

To determine where a function is increasing or decreasing, we perform an interval test on the derivative of the function. A positive derivative within an interval indicates the function is increasing there, while a negative derivative signifies a decreasing function.

For the function \(f(x) = (x-1)^{2}(x+3)\), we found the derivative helps us identify that on the interval \((-\text{\(\backslash\)}infty, -\frac{5}{3})\), the function is increasing because the derivative is positive. Conversely, on the interval \((-5/3, 1)\), the function is decreasing, given the negative value of the derivative, and then it starts increasing again on the interval \((1, \text{\(\backslash\)}infty)\). This alternating pattern of increasing and decreasing behavior is crucial in locating relative extrema.

Relative Extrema

Identifying relative extrema involves using the information gleaned from critical numbers and the intervals of increase or decrease. Relative extrema are the peaks and valleys—local maxima and minima—of the function’s graph.

Using the first derivative test, we examine the critical numbers within the context of the intervals. Where the derivative changes from positive to negative, we find a relative maximum. Where it changes from negative to positive, a relative minimum is found. For our function, at \(x = -\frac{5}{3}\), the function shows a transition from increasing to decreasing, indicating a relative maximum. At \(x = 1\), there is a shift from decreasing to increasing, revealing a relative minimum. These points are significant in understanding the overall shape and turning points of the function's graph.

Derivative of a Function

The derivative of a function is the foundation of calculating critical numbers and determining intervals of increase and decrease. It represents the function's instantaneous rate of change or slope at a given point.

It is symbolic of the steepness or incline of the graph at any particular point. For the function at hand, \(f(x) = (x-1)^{2}(x+3)\), the derivative \(f'(x) = 3x^2 - 2x - 5\) is crucial for all subsequent analysis including identifying critical points, and assessing whether the function is increasing or decreasing at different intervals. The derivative's sign helps decipher the function’s behavior—it's a great tool for predicting and explaining the function's graphic representation.