Problem 25 - Precalculus: Functions and Graphs

Question

Find all values of \(x\) such that \(f(x)>0\) and all \(x\) such that \(f(x)<0,\) and sketch the graph of \(f\) $$f(x)=x^{4}-6 x^{2}+8$$

Step-by-Step Solution

Answer

\( f(x) > 0 \) on \((-\sqrt{3}, 0)\), \((0, \sqrt{3})\); \( f(x) < 0 \) on \((-\infty, -\sqrt{3})\), \((\sqrt{3}, \infty)\).

Step 1: Identify Critical Points

Find the derivative of the function to determine where it changes sign. The function is given as \( f(x) = x^4 - 6x^2 + 8 \). First, find the first derivative: \( f'(x) = 4x^3 - 12x \). Set the derivative equal to zero: \( 4x^3 - 12x = 0 \). Factor the equation: \( 4x(x^2 - 3) = 0 \). Solve for \( x \), giving the critical points: \( x = 0, \pm \sqrt{3} \).

Step 2: Test Intervals Between Critical Points

To determine where the function \( f(x) \) is positive or negative, test the intervals: \((-\infty, -\sqrt{3})\), \((-\sqrt{3}, 0)\), \((0, \sqrt{3})\), \((\sqrt{3}, \infty)\). Choose test points from each interval and substitute them into \( f(x) \) to see where the function is positive or negative. For example, use \( x = -2 \), \( x = -1 \), \( x = 1 \), and \( x = 2 \).

Step 3: Evaluate Test Points

Evaluate at \( x = -2 \): \( f(-2) = 16 - 24 + 8 = 0 \). Evaluate at \( x = -1 \): \( f(-1) = 1 - 6 + 8 = 3 > 0 \). Evaluate at \( x = 1 \): \( f(1) = 1 - 6 + 8 = 3 > 0 \). Evaluate at \( x = 2 \): \( f(2) = 16 - 24 + 8 = 0 \). Use these evaluations to determine the sign of \( f(x) \) in each interval.

Step 4: Interpret Results

The function \( f(x) \) is positive on \((-\sqrt{3}, 0)\) and \((0, \sqrt{3})\). It is negative on \((-\infty, -\sqrt{3})\) and \((\sqrt{3}, \infty)\).

Step 5: Sketch the Graph of f(x)

Plot the points where \( f(x) = 0 \), i.e., the critical points \( x = 0, \pm \sqrt{3} \), and other zeros found from factorization \( x = \pm 2 \). Sketch the function, ensuring it goes upward (positive) between \( -\sqrt{3} \) and \( \sqrt{3} \) and downward (negative) outside these intervals, considering the behavior at critical points.

Key Concepts

Critical Points

Understanding critical points is crucial when analyzing polynomial inequalities. A critical point is where the derivative of a function changes its sign. This indicates a potential maximum, minimum, or inflection point of the function.

To find the critical points, first, calculate the derivative of the given function. In this exercise, the function was \( f(x) = x^4 - 6x^2 + 8 \). Taking the derivative gives us \( f'(x) = 4x^3 - 12x \).

Next, we equate the derivative to zero: \( 4x^3 - 12x = 0 \). Simplifying, we get \( 4x(x^2 - 3) = 0 \). Solving this equation gives us the critical points: \( x = 0 \) and \( x = \pm \sqrt{3} \).

Recognizing these critical points allows us to effectively analyze changes in the behavior of the function across different intervals.

Derivatives

Derivatives help us understand the changing behavior of a function. Essentially, they measure how a function's value changes as its input changes. For polynomial inequalities, derivatives are pivotal in determining where a function might shift from increasing to decreasing, or vice versa.

In this problem, we started with the function \( f(x) = x^4 - 6x^2 + 8 \) and derived its first-order derivative, \( f'(x) = 4x^3 - 12x \). By setting the derivative to zero, \( f'(x) = 0 \), we find points that are potential local maxima, minima, or points of inflection. Here, solving gives us \( x = 0, \pm \sqrt{3} \).

These critical points segment our polynomial into different intervals, which we'll then analyze to understand the function's behavior.

Interval Testing

Interval testing allows us to determine where our function \( f(x) \) is positive or negative by examining its behavior between critical points. Once you have identified the critical points \( x = 0 \) and \( x = \pm \sqrt{3} \), the next step is to test intervals around those points.

Break the x-axis into segments based on these critical values:

• \((-\infty, -\sqrt{3})\)
• \((-\sqrt{3}, 0)\)
• \((0, \sqrt{3})\)
• \((\sqrt{3}, \infty)\)

For each interval, choose a test point. A convenient choice might be values like \(-2, -1, 1, \) and \(2\) in this example. Substitute these values back into the original function \( f(x) \), and observe the sign of each result.

For instance, if substituting \( x = -1 \) into \( f(x) \) yields a positive value, the function is positive in \((-\sqrt{3}, 0)\). Repeat this process to map out where the function is positive or negative across all intervals. This intuitive method not only helps in sketching the graph but also solidifies your understanding of the function's behavior across its domain.