Question

Use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. $$ g(x)=(x+1)(x-2) $$

Short answer

Increasing on \((\frac{1}{2}, \, \infty)\), decreasing on \((-\infty, \, \frac{1}{2})\).

Step-by-step solution

Expand and Simplify the Function

First, expand the given function \[ g(x) = (x+1)(x-2) \]by distributing the terms. This gives us:\[ g(x) = x(x-2) + 1(x-2) = x^2 - 2x + x - 2 = x^2 - x - 2 \] This is the quadratic function we will work with to find where it is increasing or decreasing.

Find the Derivative

To determine where the function is increasing or decreasing, we need to find its first derivative. The first derivative of \[ g(x) = x^2 - x - 2 \]is given by applying standard differentiation rules:\[ g'(x) = \frac{d}{dx}(x^2) - \frac{d}{dx}(x) - \frac{d}{dx}(2) = 2x - 1 \]

Find Critical Points

Critical points occur when the first derivative is equal to zero or undefined. We solve for \[ g'(x) = 0 \]which gives:\[ 2x - 1 = 0 \]Solving for \( x \), we get:\[ 2x = 1 \]\[ x = \frac{1}{2} \].This is our critical point.

Determine the Sign of the First Derivative

We use the critical point \( x = \frac{1}{2} \) to test intervals around it to see where the function is increasing or decreasing. The function will be increasing where the derivative is positive and decreasing where it is negative.- Test an interval less than \( \frac{1}{2} \), for example, \( x = 0 \): \[ g'(0) = 2(0) - 1 = -1 \] (negative, so the function is decreasing)- Test an interval greater than \( \frac{1}{2} \), for example, \( x = 1 \): \[ g'(1) = 2(1) - 1 = 1 \] (positive, so the function is increasing)

State the Intervals of Increase and Decrease

Based on the derivative tests, summarize where the function is increasing and decreasing:- Increasing: \((\frac{1}{2}, \, \infty)\)- Decreasing: \((-\infty, \, \frac{1}{2})\)

Key concepts

Monotonicity Theorem

To understand when a function is increasing or decreasing, we can use the Monotonicity Theorem. This theorem helps us analyze the behavior of a function by examining its derivative. If a function's derivative is positive over an interval, that means the function is increasing on that interval. Conversely, if the derivative is negative, the function is decreasing.
This theorem is essential because it allows us to predict and understand the changes in the function's values without necessarily graphing it. By determining where the derivative changes its sign, we can easily identify intervals of increase and decrease. In practice, we often find critical points first and use them to segment the domain into intervals for analysis.

Critical Points

Critical points are essential in understanding where a function will change from increasing to decreasing, or vice versa. They occur at values of the domain where the derivative of the function is zero or undefined. When we solve for the critical points, as in the exercise, we set the first derivative equal to zero and solve for the variable.
For the function \[ g(x) = x^2 - x - 2 \]we found the first derivative to be \[ g'(x) = 2x - 1 \].Setting \[ g'(x) = 0 \]gives us the critical point \[ x = \frac{1}{2} \].This critical point divides the function's domain into parts where we can individually check for increasing or decreasing behavior. Critical points are key stepping stones for performing a detailed analysis of a function’s graph.

First Derivative Test

The First Derivative Test is a systematic way to determine whether a critical point is a local maximum, local minimum, or neither. This test involves checking the sign of the derivative before and after the critical point.
Given the derivative \[ g'(x) = 2x - 1 \],we tested points around the critical point \[ x = \frac{1}{2} \]:

• For a point like \[ x = 0 \], where \[ g'(0) = -1 \] (negative), it implies the function is decreasing.
• For a point like \[ x = 1 \], where \[ g'(1) = 1 \] (positive), it implies the function is increasing.

The result of this test for our example shows that the function transitions from decreasing to increasing at \[ x = \frac{1}{2} \].Therefore, in this context, \[ x = \frac{1}{2} \]is a relative minimum. This method provides a clear strategy to comprehend how a function behaves around its critical points.