Question

Where is the function \(f(x)=x^{2}-4 x+3\) increasing? Where is it decreasing?

Short answer

Increasing on (2, \infty) and decreasing on (-\infty, 2).

Step-by-step solution

- Find the Derivative

To determine where the function is increasing or decreasing, first find its derivative. The derivative of the function \(f(x) = x^2 - 4x + 3\) is \(f'(x) = 2x - 4\).

- Set the Derivative to Zero

Find the critical points by setting the derivative equal to zero: \(2x - 4 = 0\). Solve for \(x\) to get \(x = 2\).

- Test Intervals Around the Critical Point

Test intervals around the critical point to determine the sign of the derivative. Check intervals \((-\infty, 2)\) and \((2, \infty)\):For \(x < 2\), choose \ x = 1 \: \(f'(1) = 2(1) - 4 = -2\) (negative)For \(x > 2\), choose \ x = 3 \: \(f'(3) = 2(3) - 4 = 2\) (positive)

- Determine Where the Function is Increasing or Decreasing

Since \(f'(x) < 0\) for \(x < 2\), the function is decreasing on \((-\infty, 2)\). Since \(f'(x) > 0\) for \(x > 2\), the function is increasing on \((2, \infty)\).

Key concepts

derivative

To determine where a function is increasing or decreasing, we first need to find its derivative. The derivative gives us the rate of change of the function. For the function \( f(x) = x^2 - 4x + 3 \), we find the derivative by applying basic differentiation rules. Differentiating each term, we get: \( f'(x) = 2x - 4 \). This derivative will guide us in understanding the behavior of our function on different intervals.

critical points

Next, we identify the critical points of the function. Critical points occur where the derivative is zero or undefined. These points help us understand where the function's behavior changes. For our function, we set the derivative to zero: \( 2x - 4 = 0 \). Solving this equation, we find the critical point: \( x = 2 \). This point is crucial as it gives a potential change in the function's increase or decrease behavior.

interval testing

To determine the behavior around the critical point, we perform interval testing. This involves selecting values from intervals divided by the critical points and observing the sign of the derivative. For the critical point at \( x = 2 \), we examine the intervals \((-\infty, 2)\) and \((2, \infty)\):
- For \( x < 2 \), choose \( x = 1 \): \( f'(1) = 2(1) - 4 = -2 \) (negative)
- For \( x > 2 \), choose \( x = 3 \): \( f'(3) = 2(3) - 4 = 2 \) (positive)
These calculations show us the function's increase or decrease in different intervals.

function behavior

Finally, we analyze the function behavior based on our interval testing results. Since \( f'(x) \) is negative for \( x < 2 \), the function is decreasing on the interval \((-\infty, 2)\). Since \( f'(x) \) is positive for \( x > 2 \), the function is increasing on the interval \((2, \infty)\). This helps us conclude where the function \( f(x) = x^2 - 4x + 3 \) changes its behavior from decreasing to increasing.