Problem 24 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine the open intervals on which the function is increasing, decreasing, or constant. $$f(x)=x^{2}-4 x$$

Step-by-Step Solution

Answer

The function \(f(x) = x^{2} - 4x\) is decreasing on the interval \((-\infty, 2)\) and increasing on the interval \((2, \infty)\)

Step 1: Find the Derivative

The first step is to differentiate the function \(f(x) = x^{2} - 4x\). The power rule of differentiation gives us that the derivative of \(x^2\) is \(2x\), and the derivative of \(4x\) is \(4\). So, we get: \(f'(x) = 2x - 4\).

Step 2: Find Critical Points

Let's find the critical points where the derivative is zero or doesn't exist. Since a polynomial and its derivative are always defined for all real numbers, we can find the zeros of the derivative by setting \(f'(x) = 2x - 4 = 0\). Solving for \(x\) gives \(x=2\). So 2 is the single critical point.

Step 3: Determine Test Intervals

Use the critical point to divide the number line into intervals. Based on the critical point found, we have two intervals of \(x\) to test: \((-\infty, 2)\) and \((2, \infty)\).

Step 4: Test Intervals

Take a test point in each interval and evaluate the sign of the derivative at these points. Any number less than 2 can work for the first interval, so let's use 0. Plugging 0 into \(f'(x)\) gives \(-4\), which is negative, meaning the function is decreasing on the interval \((-\infty, 2)\). For the second interval, we can use 3 as a test point. Plugging in 3 gives \(2\), which is positive, so the function is increasing on the interval \((2, \infty)\). Thus, the function is decreasing from \(-\infty\) to 2 and increasing from 2 to \(\infty\).

Step 5: Check for Constant Interval

Finally, check if the function is constant on any interval by determining if there exists an interval where the derivative doesn't change signs. This is not the case here as the derivative changes sign at \(x=2\). So, there are no constant intervals.