Problem 62 - Algebra for College Students

Question

Find the intervals on which the given function is increasing and the intervals on which it is decreasing. $$ f(x)=(x-3)^{2}+1 $$

Step-by-Step Solution

Answer

Decreasing on \((-\infty, 3)\), increasing on \((3, \infty)\).

Step 1: Find the derivative

To determine where the function is increasing or decreasing, we first need to find its derivative. The given function is \( f(x) = (x-3)^2 + 1 \). Using the power rule, the derivative is \( f'(x) = 2(x-3) \).

Step 2: Identify critical points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero and solve for \( x \): \( 2(x-3) = 0 \). Solving this gives \( x = 3 \). There are no values that make the derivative undefined, so the only critical point is \( x = 3 \).

Step 3: Determine intervals

To determine where the function is increasing or decreasing, we analyze the sign of \( f'(x) \) on the intervals created by the critical point. These intervals are \((-\infty, 3)\) and \((3, \infty)\).

Step 4: Test intervals

Choose a test point from each interval to determine the sign of \( f'(x) \): - For \( x = 2 \) in the interval \((-\infty, 3)\), \( f'(2) = 2(2-3) = -2 \) (negative, so the function is decreasing).- For \( x = 4 \) in the interval \((3, \infty)\), \( f'(4) = 2(4-3) = 2 \) (positive, so the function is increasing).

Step 5: Write the conclusion

Based on our test points, the function is decreasing on \((-\infty, 3)\) and increasing on \((3, \infty)\).

Key Concepts

Critical Points

Critical points are important in understanding the behavior of functions. They are the values of \( x \) where a function might change its increasing or decreasing direction. To find critical points, we examine the derivative of the function. These points occur where the derivative is zero or undefined.

In our example, we have the function \( f(x) = (x-3)^2 + 1 \). The derivative, \( f'(x) = 2(x-3) \), is a linear function. By setting the derivative equal to zero:

• \( 2(x-3) = 0 \)
• This simplifies to \( x = 3 \)

The expression is not undefined anywhere, so \( x = 3 \) is the only critical point. This critical point helps us understand where the function changes its behavior from decreasing to increasing.

Derivative

A derivative is a tool used in calculus to understand how a function changes at any point. It reflects the rate of change or the slope of the function. For any given function \( f(x) \), its derivative \( f'(x) \) is found by applying rules from calculus such as the power rule, product rule, etc.

In our exercise, where the function is \( f(x) = (x-3)^2 + 1 \), we apply the power rule to find its derivative. The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). Therefore, the derivative is:

• \( f'(x) = 2(x-3) \)

The derivative helps us find critical points and analyze the increasing or decreasing nature of the function.

Function Analysis

Function analysis is a process to understand how a function behaves throughout its domain. This involves examining intervals where the function is increasing or decreasing. The derivative plays a central role in this analysis. By evaluating the sign of the derivative, we can determine the behavior of the function.

In our solution, we discovered the critical point \( x = 3 \). We then divided the number line into intervals based on this point: \((-\infty, 3)\) and \((3, \infty)\). By choosing test points in each interval, such as \( x = 2 \) and \( x = 4 \), we evaluate the derivative:

• \( f'(2) = 2(2-3) = -2 \), indicating a negative slope and a decreasing function
• \( f'(4) = 2(4-3) = 2 \), indicating a positive slope and an increasing function

Thus, the function is confirmed to be decreasing on \((-\infty, 3)\) and increasing on \((3, \infty)\). This careful analysis helps in interpreting the behavior of the function at different intervals.