Problem 2 - Calculus

Question

In Problems \(I-10\), 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

The function decreases on \((- fty, \frac{1}{2})\) and increases on \((\frac{1}{2}, \infty)\).

Step 1: Differentiate the Function

To find where the function is increasing or decreasing, we need to first find its derivative. Given the function \( g(x) = (x+1)(x-2) \), we apply the product rule. The derivative of \( g(x) \) is calculated as follows:\[g'(x) = \frac{d}{dx}[(x+1)(x-2)] = (x+1)'(x-2) + (x+1)(x-2)'\]\[= (1)(x-2) + (x+1)(1) = x-2 + x+1 = 2x-1\]

Step 2: Find Critical Points

Set the derivative equal to zero to find the critical points which might indicate changes in monotonicity:\[2x - 1 = 0\]Solving for \( x \), we get:\[2x = 1 \x = \frac{1}{2}\]This is the critical point where the function's behavior changes.

Step 3: Determine Intervals of Increase and Decrease

Use the critical point \( x = \frac{1}{2} \) to test intervals around it. Choose a test point from each interval: - For \( x < \frac{1}{2} \), choose \( x = 0 \): \[ g'(0) = 2(0) - 1 = -1 \] (negative, function is decreasing).- For \( x > \frac{1}{2} \), choose \( x = 1 \): \[ g'(1) = 2(1) - 1 = 1 \] (positive, function is increasing).

Step 4: Conclusion: Describe the Monotonicity of the Function

Based on the sign of the derivative:- The function \( g(x) = (x+1)(x-2) \) is decreasing on the interval \((-\infty, \frac{1}{2})\).- The function is increasing on the interval \((\frac{1}{2}, \, \infty)\).

Key Concepts

Increasing Functions

An increasing function is one that becomes larger as the input value, \( x \), increases. This can be verified by checking that the derivative of the function, \( g'(x) \), is positive over an interval.

• When \( g'(x) > 0 \), the function is rising; i.e., for every increase in \( x \), \( g(x) \) also increases.
• This means that the slope of the tangent line to the graph of the function at any point in this interval is positive.

In our exercise, after finding and analyzing the derivative of \( g(x) = (x+1)(x-2) \), we determined that for \( x > \frac{1}{2} \), \( g'(x) = 2x-1 \) is positive. Thus, the function is increasing in the interval \( (\frac{1}{2}, \, \infty) \). This tells us that as \( x \) becomes larger than \( \frac{1}{2} \), the value of \( g(x) \) increases.

Decreasing Functions

Decreasing functions display behavior that is the opposite of increasing functions. A function decreases over an interval if its derivative is negative in that interval.

• With \( g'(x) < 0 \), the function diminishes as \( x \) increases; that is, for each increase in \( x \), \( g(x) \) decreases.
• The graph of the function at any point in this interval has a tangent line with a negative slope.

In examining \( g(x) \), we noted that for \( x < \frac{1}{2} \), \( g'(x) = 2x-1 \) is negative. Therefore, the function is decreasing on \( (-\infty, \frac{1}{2}) \). This implies that as \( x \) moves towards \( \frac{1}{2} \) from the left, the function \( g(x) \) decreases.

Derivatives

Derivatives are a core concept in calculus, they allow us to measure how a function changes as its input changes.

• A derivative, expressed as \( g'(x) \), gives us the slope of the tangent line to the curve at any point \( x \).
• Finding the derivative involves rules like the product rule, which is used when differentiating products of functions.

For our problem, to find \( g'(x) \), we applied the product rule to the function \( g(x) = (x+1)(x-2) \). The calculation was simplified to \( g'(x) = 2x - 1 \). This formula is essential for determining intervals where the function increases or decreases.

Critical Points

Critical points are where the derivative of a function is zero or undefined, indicating potential changes in the function's behavior.

• They help identify where a function shifts between increasing and decreasing or vice versa.
• These points are crucial in analyzing the function's overall shape and finding local minima or maxima.

In our case, we found a critical point at \( x = \frac{1}{2} \) by setting \( g'(x) = 2x - 1 = 0 \). Solving this gave \( x = \frac{1}{2} \). This point divides the intervals where \( g(x) \) increases or decreases and provides a deeper understanding of the function's behavior across its domain.