Problem 15 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. $$ f(x)=x^{2}-3 x $$

Step-by-Step Solution

Answer

The function is decreasing on the interval \((-\infty, \frac{3}{2})\) and increasing on the interval \((\frac{3}{2}, \infty)\).

Step 1: Find the first derivative of the function

To find the first derivative of the function \( f(x) = x^2 - 3x \), we use the power rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). So, the first derivative of \( f(x) \) is $$ f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3. $$

Step 2: Set the first derivative to zero and solve for x

To find the critical points, equate the first derivative to zero and solve for \(x\): $$ 2x - 3 = 0 $$ Solving for \(x\), we get $$ x = \frac{3}{2}. $$ This is the only critical point of the function.

Step 3: Test the intervals determined by the critical point

After finding the critical point \(x = \frac{3}{2}\), we must test the sign of the first derivative on the intervals determined by this critical point. The intervals are \(x < \frac{3}{2}\) and \(x > \frac{3}{2}\). For \(x < \frac{3}{2}\): Choose \(x = 1\). Then, $$ f'(1) = 2(1) - 3 = -1 < 0, $$ meaning that the function is decreasing on this interval. For \(x > \frac{3}{2}\): Choose \(x = 2\). Then, $$ f'(2) = 2(2) - 3 = 1 > 0, $$ meaning that the function is increasing on this interval.

Step 4: Write the final answer

The function is decreasing on the interval \((-\infty, \frac{3}{2})\) and increasing on the interval \((\frac{3}{2}, \infty)\).

Key Concepts

First Derivative Test

Understanding the behavior of a function is crucial, and that's where the first derivative test shines. It allows us to determine whether a function is increasing or decreasing at specific points. Simply put, when the first derivative of a function is positive over an interval, the function is increasing there. Conversely, if the first derivative is negative, the function decreases.

After determining critical points, we divide our number line into intervals around these points. We then select test points from each interval to evaluate the sign of the first derivative. If we find the derivative switches from negative to positive at a critical point, it indicates a local minimum; a switch from positive to negative suggests a local maximum. In our function's case, the derivative changed from negative to positive at the critical point, marking the start of an increasing interval.

Critical Points

So, what exactly are 'critical points'? These are values of 'x' where the first derivative of a function equals zero or doesn't exist. Finding them is a core step in analyzing the function's behavior. Critical points are potential locations where a function can switch direction, from increasing to decreasing or vice versa.

In our exercise, by setting the first derivative to zero, we determined the critical point at \( x = \frac{3}{2} \). This single point partitions the number line into two regions to investigate further. It's important to check for more than one critical point, as complex functions can have multiple turning points.

Power Rule

The power rule is a quick tool for differentiation, especially when dealing with polynomials. Here's the gist: if you have a function \( f(x) = x^n \), its derivative with respect to 'x' is given by \( f'(x) = nx^{n-1} \). This rule simplifies finding derivatives, cutting down on time-consuming calculations.

In our function \( f(x) = x^2 - 3x \), we applied the power rule to each term separately. The derivative of \( x^2 \) is \( 2x \) and for \( -3x \) it's simply \( -3 \). Combining these, we get the first derivative \( f'(x) = 2x - 3 \) as stated in the solution.

Function Analysis

Function analysis encompasses several techniques used to scrutinize the behavior of functions. By examining a function's first and second derivatives, we can discern intervals of increase and decrease, identify maxima and minima, and even talk about concavity and points of inflection.

With the exercise at hand, we focused on finding intervals where the function increases or decreases. Step by step, we started by finding the function's derivative, identifying critical points, and applying the first derivative test. These elements of function analysis are essential for understanding comprehensive graphs and predicting function behavior over different intervals.