Problem 27 - Algebra and Trigonometry

Question

23- 30 . A function \(f\) is given. (a) Use a graphing device to draw the graph of \(f .\) (b) State approximately the intervals on which \(f\) is increasing and on which \(f\) is decreasing. $$ f(x)=x^{3}+2 x^{2}-x-2 $$

Step-by-Step Solution

Answer

Increasing: \((-\infty, -1.55)\) and \((0.55, \infty)\). Decreasing: \((-1.55, 0.55)\).

Step 1: Graph the Function

Using a graphing calculator or online graphing tool, plot the function \( f(x) = x^3 + 2x^2 - x - 2 \). Observe the shape and direction of the curve. This will help in identifying the regions where the function is increasing or decreasing.

Step 2: Identify Critical Points

To determine where the function changes from increasing to decreasing or vice versa, find its derivative \( f'(x) \). Compute \( f'(x) = 3x^2 + 4x - 1 \). Solve the equation \( 3x^2 + 4x - 1 = 0 \) to find the critical points. Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3, b = 4, \) and \( c = -1 \). This gives \( x = \frac{-4 \pm \sqrt{16 + 12}}{6} = \frac{-4 \pm \sqrt{28}}{6} \). Approximate solutions are \( x \approx -1.55 \) and \( x \approx 0.55 \).

Step 3: Test Intervals for Increasing or Decreasing Behavior

Choose values around the critical points \( x \approx -1.55 \) and \( x \approx 0.55 \) to test if the function is increasing or decreasing. If \( f'(x) > 0 \), then \( f \) is increasing; if \( f'(x) < 0 \), then \( f \) is decreasing. For \( x < -1.55 \), test \( x = -2 \) which gives \( f'(-2) = 3(-2)^2 + 4(-2) - 1 = 7 \) (increasing). For \( -1.55 < x < 0.55 \), test \( x = 0 \), which gives \( f'(0) = -1 \) (decreasing). For \( x > 0.55 \), test \( x = 1 \), which gives \( f'(1) = 6 \) (increasing).

Step 4: State the Intervals

The function \( f(x) \) is increasing on the intervals \( (-\infty, -1.55) \) and \( (0.55, \infty) \). The function is decreasing on the interval \( (-1.55, 0.55) \).

Key Concepts

Graphing Polynomial Functions

Graphing polynomial functions involves plotting curves described by a polynomial equation. These equations are made up of terms with variable exponents that are non-negative integers. For the function given, \( f(x) = x^3 + 2x^2 - x - 2 \), we graph it using a graphing calculator or tool.

Here’s how you can approach it:

• Observe the end behavior: Because this is a cubic equation (highest degree is 3), the ends of the graph will extend in opposite directions. As \( x \) goes to infinity, \( f(x) \) also tends to infinity, while as \( x \) tends to negative infinity, \( f(x) \) will go to negative infinity.
• Identify any turning points or inflection points: These can show changes in the increasing or decreasing nature of the function as well as curvature.

The graph will give visual intuition where the function might increase or decrease, and helps verify calculations. This essential graphing step aids in understanding the real-world behavior of polynomial models.

Critical Points

Critical points are where the function's derivative, \( f'(x) \), is zero or undefined. They help identify potential maximums, minimums, or inflection points in the function.

For \( f(x) = x^3 + 2x^2 - x - 2 \), compute its derivative: \( f'(x) = 3x^2 + 4x - 1 \). Solve \( f'(x) = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 3 \), \( b = 4 \), \( c = -1 \).

• Substitute values to get critical points: \( x \approx -1.55 \) and \( x \approx 0.55 \).
• These points are where the function could switch from increasing to decreasing, or vice versa.

Checking around these critical points will confirm the behavior of the function such as if they are minimums or maximums.

Increasing and Decreasing Intervals

Understanding increasing and decreasing intervals involves analyzing the sign of the derivative \( f'(x) \). If the derivative is positive in an interval, the function is increasing in that region; if it's negative, the function decreases.

Let's apply this to our critical points for \( f(x) \):

• Choose test points near critical values: before \( -1.55 \), in between \( -1.55 \) and \( 0.55 \), and after \( 0.55 \).
• Evaluate \( f'(x) \) at these points: For \( x = -2 \), \( f'(-2) = 7 \) (positive, increasing). For \( x = 0 \), \( f'(0) = -1 \) (negative, decreasing). For \( x = 1 \), \( f'(1) = 6 \) (positive, increasing).

Thus, we determine the intervals as follows:

• Increasing on \( (-\infty, -1.55) \) and \( (0.55, \infty) \).
• Decreasing on \( (-1.55, 0.55) \).

Recognizing these intervals is a key part of understanding the function's behavior, particularly in predicting trends.