Problem 3 - Essential Calculus

Question

Identify the open intervals on which the function is increasing or decreasing. $$ f(x)=x^{2}-6 x+8 $$

Step-by-Step Solution

Answer

The function \(f(x) = x^{2} - 6x + 8\) is decreasing for \(x < 3\) and increasing for \(x > 3\).

Step 1: Find the Derivative

The derivative of the function \(f(x) = x^{2} - 6x + 8\) is given by \(f'(x) = 2x - 6\).

Step 2: Set the Derivative Equal to Zero

Setting the derivative equal to zero gives \(2x - 6 = 0\). Solving for \(x\), we have \(x = 3\).

Step 3: Determine the Sign of the Derivative

We analyze the behavior of the derivative in the intervals determined by the critical point \(x = 3\). Left of 3 (say \(x = 0\)), we see that \(f'(x) = -6 < 0\). So, the function is decreasing for \(x < 3\). Right of 3 (say \(x = 4\)), we have \(f'(x) = 2 > 0\). So, the function is increasing for \(x > 3\).

Key Concepts

Derivative

To understand whether a function is increasing or decreasing, we start by finding its derivative. The derivative of a function essentially tells us how the function's output changes as its input changes. In this case, we have the function \( f(x) = x^2 - 6x + 8 \). By differentiating, or finding the derivative, we obtain \( f'(x) = 2x - 6 \). This new expression, \( f'(x) \), gives us the slope of the tangent line at any point on the curve of \( f(x) \).

When the derivative is positive, the function is increasing because the slope of the tangent line is going up. Conversely, a negative derivative indicates a decreasing function. So analyzing \( f'(x) \) helps us understand how the original function behaves as \( x \) changes.

Increasing and Decreasing Functions

With the derivative in hand, we explore intervals where the function is increasing or decreasing. We do this by setting the derivative \( f'(x) = 2x - 6 \) to zero to find critical points. Critical points are values of \( x \) where the slope changes, which are potential "turning points" for our function.

Here, the critical point is \( x = 3 \), as \( 2x - 6 = 0 \) simplifies to \( x = 3 \). Now, let's examine the intervals around this critical point:

• For \( x < 3 \): Picking a test point like \( x = 0 \), we find \( f'(0) = -6 \), implying a negative slope, so \( f(x) \) is decreasing.
• For \( x > 3 \): Try \( x = 4 \) to calculate \( f'(4) = 2 \), showing a positive slope, so \( f(x) \) is increasing.

These analyses show us that the function decreases on the interval \((-\infty, 3)\) and increases on \((3, \infty)\). But why is this important? Determining these intervals helps us understand the graph's shape without having to plot points exhaustively.

Critical Points

Critical points are vital in understanding a function's behavior. They occur where the derivative is zero or undefined—places where the function might stop increasing or decreasing. In our exercise, we discovered a critical point at \( x = 3 \).

Finding critical points is the first step to sketching a function's graph or solving optimization problems. They signal where the function could switch from increasing to decreasing or vice versa. However, just identifying a critical point doesn't tell us everything. We use it as a starting point to test intervals around it, as demonstrated earlier.

• The test confirmed intervals \(-\infty < x < 3\), where the slope was negative, indicating decrease.
• For \( x > 3 \), the slope turned positive, indicating increase.

Always remember, finding and testing around critical points is essential in calculus for understanding and describing the behavior of functions efficiently.