Problem 22 - 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. $$ f(x)=x^{3}-6 x^{2}+15 $$

Step-by-Step Solution

Answer

The critical numbers of the function are \(0\) and \(4\). The function is increasing on the intervals \(-\infty,0\) and \(4,\infty\), and decreasing on the interval \(0,4\). At \(x=0\), \(f(x)\) has a maximum at \(15\), and at \(x=4\), \(f(x)\) has a minimum at \(4\).

Step 1: Find the derivative of the function

Differentiate the function \(f(x)=x^{3}-6x^{2}+15\). The derivative \(f'(x)\) can be found using the power rule: Given \(f(x) = ax^n\), the derivative is \(f'(x)=nax^{n-1}\). Applying this rule to all terms in the function results in \(f'(x)=3x^{2}-12x\).

Step 2: Find the critical numbers

Critical numbers are the numbers that make the derivative equal to zero or undefined. However, in this case \(f'(x)=3x^{2}-12x\) never gets undefined. Thus, we solve for \(x\) in equation \(f'(x)=0\). This gives us \(0=3x^{2}-12x\), and by factoring out a common factor of \(3x\), we get \(x=0,4\). These are the critical numbers.

Step 3: Determine the intervals where the function is increasing or decreasing

The First Derivative Test is used to determine where the function increases or decreases. It states that if \(f'(x)\) changes from positive to negative at a point, then \(f(x)\) has a local maximum at that point. Conversely, if \(f'(x)\) changes from negative to positive, then \(f(x)\) has a local minimum. Plotting and testing the intervals around the critical numbers \(0\) and \(4\) in the derivative equation shows that \(f'(x)\) is positive for \(x<0\) and \(x>4\) and negative for \(0

Step 4: Identify the relative extrema

The extrema are defined at the critical numbers. The change in the sign of \(f'(x)\) can be used to identify them. At \(x=0\), \(f'(x)\) changes from positive to negative, which means that \(f(0)\) is a local maximum. Conversely, at \(x=4\), \(f'(x)\) changes from negative to positive, indicating a local minimum. Thus, the relative extrema are identified, where \(f(0)=15\) is the local maximum and \(f(4)=4\) is the local minimum.

Step 5: Confirm the results graphically

Plot the function \(f(x)=x^{3}-6x^{2}+15\) to confirm the results. The function should show a local maximum at \(x=0\) and a local minimum at \(x=4\), and it should be increasing for \(x<0\) and \(x>4\) and decreasing for \(0

Key Concepts

Critical Numbers

Understanding critical numbers is crucial when analyzing the behavior of functions. These numbers refer to points on the graph of a function where the derivative is zero or undefined. They are essential in finding local maxima and minima, which are high or low points relative to their surroundings. In the given exercise, critical numbers are found by setting the derivative of the function equal to zero. Since the derivative expressed as f'(x) = 3x^2 - 12x is a polynomial, it won’t be undefined for any real number. By factoring, we find two critical numbers: x = 0 and x = 4. These are the points we'll further analyze to determine if they correspond to relative extrema.

Increasing and Decreasing Intervals

To understand how a function behaves, we study the intervals where it is increasing or decreasing. A function is 'increasing' if its slope is positive, signaling that as x gets larger, so does f(x). Conversely, it's 'decreasing' where the slope is negative. Here, we use the derivative to determine these intervals. By testing values around the critical numbers, we see that f'(x) is positive for x < 0 and x > 4, indicating the function is increasing there. In between, for 0 < x < 4, the derivative is negative, showing the function is decreasing. Therefore, these intervals guide us to predict the function's behavior without even looking at its graph.

Relative Extrema

Relative extrema refer to the local highs (maxima) and lows (minima) on a function's graph. The First Derivative Test helps us locate these points using critical numbers. If the derivative changes from positive to negative at a critical number, we have a local maximum. Conversely, if the derivative changes from negative to positive, that point is a local minimum. For the function f(x) = x^3 - 6x^2 + 15, the critical numbers x = 0 and x = 4 indicate a local maximum and minimum, respectively. The test shows f'(x) changes from positive to negative at zero and negative to positive at four, confirming these are indeed relative extrema.

Graphical Analysis

A graphical representation of the function can confirm what we've determined algebraically. By plotting f(x) = x^3 - 6x^2 + 15 on a graphing utility, we should observe the behavior predicted by the previous steps: an increase, followed by a decrease, and then another increase in the function's values. Specifically, the graph will display a peak at the local maximum x = 0 and a valley at the local minimum x = 4. This visual cue not only confirms the calculations of increasing and decreasing intervals and relative extrema but also provides a way to verify our complete understanding of the function's behavior over its domain.

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Problem 22

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Problem 23

Other exercises in this chapter

Problem 22

Finding Points of Inflection In Exercises \(15-30\) , find the points of inflection and discuss the concavity of the graph of the function. $$ f(x)=x \sqrt{9-x}

View solution

Problem 22

Finding Extrema on a Closed Interval In Exercises \(17-36\) , find the absolute extrema of the function on the closed interval. $$ f(x)=2 x^{3}-6 x,[0,3] $$

View solution

Problem 23

Finding a Limit In Exercises \(19-38,\) find the limit. $$ \lim _{x \rightarrow \infty} \frac{x}{x^{2}-1} $$

View solution

Problem 23

Approximate the fixed point of the function to two decimal places. [A fixed point \(x_{0}\) of a function \(f\) is a value of \(x\) such that \(f\left(x_{0}\rig

View solution

Key Concepts

Critical Numbers

Understanding critical numbers is crucial when analyzing the behavior of functions. These numbers refer to points on the graph of a function where the derivative is zero or undefined. They are essential in finding local maxima and minima, which are high or low points relative to their surroundings. In the given exercise, critical numbers are found by setting the derivative of the function equal to zero. Since the derivative expressed as f'(x) = 3x^2 - 12x is a polynomial, it won’t be undefined for any real number. By factoring, we find two critical numbers: x = 0 and x = 4. These are the points we'll further analyze to determine if they correspond to relative extrema.

Increasing and Decreasing Intervals

To understand how a function behaves, we study the intervals where it is increasing or decreasing. A function is 'increasing' if its slope is positive, signaling that as x gets larger, so does f(x). Conversely, it's 'decreasing' where the slope is negative. Here, we use the derivative to determine these intervals. By testing values around the critical numbers, we see that f'(x) is positive for x < 0 and x > 4, indicating the function is increasing there. In between, for 0 < x < 4, the derivative is negative, showing the function is decreasing. Therefore, these intervals guide us to predict the function's behavior without even looking at its graph.

Relative Extrema

Relative extrema refer to the local highs (maxima) and lows (minima) on a function's graph. The First Derivative Test helps us locate these points using critical numbers. If the derivative changes from positive to negative at a critical number, we have a local maximum. Conversely, if the derivative changes from negative to positive, that point is a local minimum. For the function f(x) = x^3 - 6x^2 + 15, the critical numbers x = 0 and x = 4 indicate a local maximum and minimum, respectively. The test shows f'(x) changes from positive to negative at zero and negative to positive at four, confirming these are indeed relative extrema.

Graphical Analysis

A graphical representation of the function can confirm what we've determined algebraically. By plotting f(x) = x^3 - 6x^2 + 15 on a graphing utility, we should observe the behavior predicted by the previous steps: an increase, followed by a decrease, and then another increase in the function's values. Specifically, the graph will display a peak at the local maximum x = 0 and a valley at the local minimum x = 4. This visual cue not only confirms the calculations of increasing and decreasing intervals and relative extrema but also provides a way to verify our complete understanding of the function's behavior over its domain.