Problem 21 - Calculus Early Transcendentals

Question

Find the intervals on which \(f\) is increasing and decreasing. Superimpose the graphs of \(f\) and \(f^{\prime}\) to verify your work. $$f(x)=12+x-x^{2}$$

Step-by-Step Solution

Answer

Answer: The function is increasing on the interval \((-\infty, \frac{1}{2})\) and decreasing on the interval \((\frac{1}{2}, \infty)\).

Step 1: Finding the derivative of the function

To find the derivative of the function \(f(x) = 12 + x - x^2\), we differentiate each term with respect to \(x\), using the power rule for derivatives: $$f'(x) = \frac{d}{dx}(12) + \frac{d}{dx}(x) - \frac{d}{dx}(x^2)$$ That gives us: $$f'(x) = 0 + 1 - 2x$$ And simplifying it yields: $$f'(x) = 1 - 2x$$

Step 2: Finding the critical points

To find the critical points, we set \(f'(x)\) equal to 0 and solve for \(x\): $$1 - 2x = 0$$ Adding \(2x\) to both sides and dividing by \(2\), we get: $$x = \frac{1}{2}$$ So, there is one critical point: \(x=\frac{1}{2}\).

Step 3: Determining intervals of increasing and decreasing

Using the critical point \(x=\frac{1}{2}\), we divide the x-axis into two intervals: \((-\infty, \frac{1}{2})\) and \((\frac{1}{2}, \infty)\). Next, we will analyze the sign of \(f'(x)\) within each interval. Pick a test point within the interval, and evaluate the sign of \(f'(x)\) at that test point. For the interval \((-\infty, \frac{1}{2})\), let's test \(x = 0\): $$f'(0) = 1 - 2(0) = 1$$ Since \(f'(x)\) is positive within this interval, the function \(f(x)\) is increasing in \((-\infty, \frac{1}{2})\). For the interval \((\frac{1}{2}, \infty)\), let's test \(x = 1\): $$f'(1) = 1 - 2(1) = -1$$ Since \(f'(x)\) is negative within this interval, the function \(f(x)\) is decreasing in \((\frac{1}{2}, \infty)\).

Step 4: Superimposing the graphs of \(f(x)\) and \(f'(x)\)

Once you have sketched the graph of both \(f(x) = 12 + x - x^2\) and \(f'(x) = 1 - 2x\) (you can use a graphing calculator or online graphing tool), verify that the behavior of \(f(x)\) corresponds with the sign of \(f'(x)\) in each interval. The function should be increasing where \(f'(x)\) is positive and decreasing where \(f'(x)\) is negative. In conclusion, the function \(f(x) = 12 + x - x^2\) is increasing on \((-\infty, \frac{1}{2})\) and decreasing on \((\frac{1}{2}, \infty)\).

Key Concepts

Derivative

In calculus, a derivative is a fundamental concept that measures how a function changes as its input changes. It finds the rate at which a quantity changes and can be interpreted as the slope of the function's graph at a given point. To determine the derivative of the function \(f(x) = 12 + x - x^2\), we applied the power rule:

• Constant terms, like 12, have a derivative of 0 because their rate of change is zero.
• For the term \(x\), its derivative is 1 because it's a linear term with a slope of 1.
• The term \(-x^2\) uses the power rule, where \(n\) is the exponent: the derivative is \(-2x\) (multiply the exponent by the coefficient and reduce the exponent by one).

Combine these derivatives, and we find \(f'(x) = 1 - 2x\). The derivative \(f'(x)\) tells us the slope of the curve at any \(x\)-value, and it's essential for finding critical points and analyzing the function's behavior.

Critical Points

Critical points occur where a function's derivative equals zero or is undefined. These points are where the function might have a local maximum or minimum, or change its increasing or decreasing behavior. For our function \(f(x) = 12 + x - x^2\), the critical point helps us explore potential changes in the graph's direction.To find the critical point, we set the derivative \(f'(x) = 1 - 2x\) equal to zero and solve for \(x\):

• Set \(1 - 2x = 0\).
• Solve for \(x\) by adding \(2x\) to both sides, yielding \(1 = 2x\).
• Divide by 2, resulting in \(x = \frac{1}{2}\).

Thus, the critical point is at \(x = \frac{1}{2}\). At this \(x\)-value, the function's slope is zero, indicating a potential shift from increasing to decreasing behavior on the graph. It marks the turning point of the curve.

Increasing and Decreasing Intervals

Identifying where a function is increasing or decreasing involves analyzing the sign of its derivative on each side of critical points.With the critical point at \(x = \frac{1}{2}\), we explore the function \(f(x) = 12 + x - x^2\) on the intervals \((-\infty, \frac{1}{2})\) and \((\frac{1}{2}, \infty)\):For \((-\infty, \frac{1}{2})\):

• Choose \(x = 0\) as a test point.
• Calculate \(f'(0) = 1 - 2(0) = 1\), which is positive.

The positive derivative indicates the function is increasing on this interval.For \((\frac{1}{2}, \infty)\):

• Choose \(x = 1\) as a test point.
• Calculate \(f'(1) = 1 - 2(1) = -1\), which is negative.

The negative derivative indicates the function is decreasing on this interval.In summary, understanding where a function is increasing or decreasing helps visualize its general behavior and curve movement. This method aids in sketching graphs and predicting function trends.