Problem 4 - University Calculus: Early Transcendentals

Question

Answer the following questions about the functions whose derivatives are given. a. What are the critical points of \(f ?\) b. On what open intervals is \(f\) increasing or decreasing? c. At what points, if any, does \(f\) assume local maximum and minimum values? $$f^{\prime}(x)=(x-1)^{2}(x+2)^{2}$$

Step-by-Step Solution

Answer

Critical points: \( x = 1 \) and \( x = -2 \). Function is always increasing; no local maxima or minima.

Step 1: Find Critical Points

To find the critical points of the function \( f \), we set its derivative \( f'(x) \) equal to zero. Since \( f'(x) = (x-1)^2(x+2)^2 \), the critical points are the solutions to \[ (x-1)^2(x+2)^2 = 0. \]This equation is solved when either \((x-1)^2 = 0\) or \((x+2)^2 = 0\). Thus, the critical points are \( x=1 \) and \( x=-2 \).

Step 2: Determine Intervals of Increase/Decrease

To determine where \( f \) is increasing or decreasing, analyze the sign of \( f'(x) \) on intervals divided by the critical points. Choose test points: - For interval \((-\infty, -2)\), choose \( x = -3 \): \((x-1)^2(x+2)^2 > 0\) because both squared terms are positive.- For interval \((-2, 1)\), choose \( x = 0 \): \((x-1)^2(x+2)^2 > 0\) again since both squared expressions are positive.- For interval \((1, \infty)\), choose \( x = 2 \): \((x-1)^2(x+2)^2 > 0\) as both terms remain positive.Since \( f'(x) \) is non-negative on all intervals and never negative, \( f \) does not decrease on any interval.

Step 3: Identify Local Maxima and Minima

Since \( f'(x) > 0 \) for all points in each interval, \( f \) maintains increasing behavior, without turning from increasing to decreasing or vice versa. Therefore, there are no local maxima or minima at the critical points \( x = 1 \) or \( x = -2 \). The function does not switch from positive to negative or negative to positive, excluding local extremum possibilities.

Key Concepts

Increasing and Decreasing Intervals

In calculus, determining whether a function is increasing or decreasing on certain intervals can be a crucial aspect of understanding the function's behavior. When we have the derivative of a function, denoted as \( f'(x) \), it tells us how the function \( f(x) \) is changing at every point.

For the function in our exercise, \( f'(x) = (x-1)^2(x+2)^2 \), we're interested in the sign of \( f'(x) \) on different intervals. The critical points we found are \( x = 1 \) and \( x = -2 \), which divide our number line into intervals: \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\).

By choosing test points in each of these intervals:

• In the interval \((-\infty, -2)\), pick \( x = -3 \). Here, both terms \((x-1)^2\) and \((x+2)^2\) are positive, making \( f'(x) > 0 \).
• In the interval \((-2, 1)\), choose \( x = 0 \). Both squared terms are positive again, so \( f'(x) > 0 \).
• In the interval \((1, \infty)\), select \( x = 2 \). Similarly, both squared expressions are positive, leading to \( f'(x) > 0 \).

Because \( f'(x) \) is always non-negative in these intervals, the function \( f(x) \) is increasing across these intervals. There is no interval where \( f(x) \) is decreasing, as the derivative never takes a negative value.

Local Maxima

A local maximum is a point in the domain of a function where the function reaches a peak value, appearing like the top of a hill in a graph. Typically, this occurs where the function changes from increasing to decreasing.

In our function, we observe that the derivative \( f'(x) = (x-1)^2(x+2)^2 \) is always positive for any test point selected in the intervals defined by our critical points. The derivative does not change sign, indicating that \( f(x) \) doesn't transition from increasing to decreasing, which would be required for a local maximum to exist.

Since \( f(x) \) is consistently increasing and never decreases at any point, it does not achieve a peak or local maximum at any point, including the critical points \( x = 1 \) and \( x = -2 \). The absence of a change in the direction of \( f(x) \) means there's no place where \( f(x) \) reaches a local highest point.

Local Minima

Similar to a local maximum, a local minimum is a point where a function reaches its lowest value in a small neighborhood or interval around that point, resembling the bottom of a valley on a graph.

Typically, a local minimum occurs when a function changes from decreasing to increasing, marked by a shift in the sign of the derivative from negative to positive. However, in our function's derivative \( f'(x) = (x-1)^2(x+2)^2 \), the derivative is consistently positive. There is no change in sign.– it's positive regardless of the interval you check, which means \( f(x) \) is entirely in an increasing trend and does not form a valley or local minimum.

Consequently, since \( f(x) \) continuously increases without any intervals of decrease, it does not display any local minima at the critical points \( x = 1 \) or \( x = -2 \). The function doesn't shift its behavior to have a downward slide, hence no local minima can be identified.