Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-10, use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. \(g(x)=(x+1)(x-2)\)

Step-by-Step Solution

Answer

\( g(x) \) is decreasing on \((- \infty, \frac{1}{2})\) and increasing on \((\frac{1}{2}, \infty)\).

Step 1: Expand the Function

First, let's expand the given function. The function provided is \[ g(x) = (x + 1)(x - 2) \]To expand this, use the distributive property (also known as the FOIL method):\[ g(x) = x^2 - 2x + x - 2 \]which simplifies to:\[ g(x) = x^2 - x - 2 \]

Step 2: Find the Derivative

To apply the Monotonicity Theorem, we need to find the derivative of the function. The derivative of our function, \[ g'(x) = \frac{d}{dx}(x^2 - x - 2) \], is calculated as:\[ g'(x) = 2x - 1 \]

Step 3: Set the Derivative Equal to Zero

We find critical points by setting the derivative equal to zero:\[ 2x - 1 = 0 \]Solving for \( x \), we get:\[ 2x = 1 \]\[ x = \frac{1}{2} \]

Step 4: Determine Intervals on the Number Line

The critical point \( x = \frac{1}{2} \) divides the number line into two intervals: 1. \((-\infty, \frac{1}{2})\)2. \((\frac{1}{2}, \infty)\)We need to test the sign of the derivative in each interval.

Step 5: Test for Increasing or Decreasing

Choose a test point from each interval and substitute it into the derivative \( g'(x) = 2x - 1 \):- For the interval \((-\infty, \frac{1}{2})\), pick \( x = 0 \):\[ g'(0) = 2(0) - 1 = -1 \] (Negative, so \( g(x) \) is decreasing here.)- For the interval \((\frac{1}{2}, \infty)\), pick \( x = 1 \):\[ g'(1) = 2(1) - 1 = 1 \] (Positive, so \( g(x) \) is increasing here.)

Key Concepts

Derivative

When dealing with functions and their behavior, the derivative is a powerful tool. It tells us how a function changes at any given point. Essentially, the derivative of a function at a particular point gives the slope of the tangent line to the function at that point, which indicates whether the function is increasing, decreasing, or constant. To find the derivative, you take the function and apply rules such as the power rule, product rule, or chain rule depending on the form of the function.

For our function, after expanding it, we arrived at a polynomial form:

• The function: \( g(x) = x^2 - x - 2 \)
• Its derivative: \( g'(x) = 2x - 1 \)

This derivative shows us that the rate of change in \( g(x) \) is linear, as indicated by the term \( 2x \). By analyzing this derivative, we can determine where the function is increasing or decreasing.

Function Expansion

Function expansion involves rewriting a function that is presented in a compact form, like a product of binomials, into a more extended, usable form such as a polynomial. For the function \( g(x) = (x + 1)(x - 2) \), expanding through the distributive property helps simplify analysis.

Here's how you can expand:

• Apply the distributive property (FOIL - First, Outer, Inner, Last): \( (x+1)(x-2) = x\times x + x\times (-2) + 1\times x + 1\times (-2) \).
• Simplify the expression to get: \( x^2 - x - 2 \).

Once expanded, it's easier to differentiate, because each term is straightforward to handle. This polynomial form reveals the roots and behavior necessary for further calculus operations like finding derivatives and critical points.

Critical Points

A critical point of a function is where the derivative is zero or undefined. These points are essential as they often indicate potential maxima, minima, or points of inflection.

To find critical points for the function \( g(x) \):

• Start with the derivative: \( g'(x) = 2x - 1 \).
• Set the derivative equal to zero: \( 2x - 1 = 0 \).
• Solve for \( x \): \( x = \frac{1}{2} \).

So, \( x = \frac{1}{2} \) is a critical point of this function. At this point, the function's rate of change shifts, highlighting a transition from increasing to decreasing behavior or vice versa.

Intervals on Number Line

Once you have determined the critical points, you can split the number line into sections to analyze the behavior of the function in each interval. In this case, the critical point divides the number line into:

• Interval 1: \(( -\infty, \frac{1}{2}) \)
• Interval 2: \(( \frac{1}{2}, \infty ) \)

Testing the derivative within these intervals gives insight into where \( g(x) \) is increasing or decreasing.

• For \(( -\infty, \frac{1}{2}) \), select a point like \( x = 0 \):
  \( g'(0) = -1 \), thus \( g(x) \) is decreasing.
• For \(( \frac{1}{2}, \infty ) \), select \( x = 1 \):
  \( g'(1) = 1 \), hence \( g(x) \) is increasing.

By determining the sign of the derivative in each interval, you correctly identify where the function is increasing or decreasing.