Problem 10 - Algebra and Trigonometry

Question

\(5-12\) . 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. $$ f(x)=x^{4}-16 x^{2} $$

Step-by-Step Solution

Answer

The function increases on \((-\infty, -\\\sqrt{8})\) and \((\\\sqrt{8}, \infty)\); decreases on \((-\\\sqrt{8}, 0)\) and \((0, \\\sqrt{8})\).

Step 1: Sketch the Graph of the Function

To graph the function \(f(x) = x^4 - 16x^2\), we use a graphing calculator or software. Start by identifying key features like intercepts and symmetries. Notice that it is a polynomial, and thus, it will have end behaviors that rise to positive infinity as \(x\) goes to infinity or negative infinity.

Step 2: Identify Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, we need to find the derivative \(f'(x)\). Calculate \(f'(x) = 4x^3 - 32x\). Set \(f'(x) = 0\) to find critical points: \(4x(x^2 - 8) = 0\), leading to critical points at \(x = 0, x = \pm \sqrt{8}\). Evaluate \(f'(x)\) in intervals created by these critical points.

Step 3: Test Sign of Derivative and Determine Behavior

In each interval \((-\infty, -\sqrt{8})\), \(-\sqrt{8}, 0\), \(0, \sqrt{8})\), and \(\sqrt{8}, \infty)\), test points like \(x = -3, -2, 1, 3\) in \(f'(x)\):- For \(x < -\sqrt{8}\), \(f'(x) > 0\) (function is increasing).- For \(-\sqrt{8} < x < 0\), \(f'(x) < 0\) (function is decreasing).- For \(0 < x < \sqrt{8}\), \(f'(x) < 0\) (function is decreasing).- For \(x > \sqrt{8}\), \(f'(x) > 0\) (function is increasing).

Step 4: Conclude with Intervals

Based on the sign of the derivative and the tested intervals, the function \(f(x)\) is increasing on \((-\infty, -\sqrt{8})\) and \((\sqrt{8}, \infty)\), and decreasing on \((-\sqrt{8}, 0)\) and \((0, \sqrt{8})\).

Key Concepts

Derivatives

Derivatives are a fundamental tool in calculus used to determine how a function changes. For a polynomial function like \(f(x) = x^4 - 16x^2\), the first derivative \(f'(x)\) gives us valuable insights about the function's behavior. To find \(f'(x)\), apply basic differentiation rules to get \(f'(x) = 4x^3 - 32x\). This derivative helps us identify where our function is increasing or decreasing by showing the slope of the tangent line at any point \(x\). A positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function. We'll utilize this concept later to explore how the derivative relates to critical points and intervals of increase and decrease.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. They are key in finding where a function changes direction. For \(f(x) = x^4 - 16x^2\), setting \(f'(x) = 0\) gives us the equation \(4x(x^2 - 8) = 0\). By solving this, we find the critical points at \(x = 0, x = \pm \sqrt{8}\). These points divide the number line into intervals, each potentially showing different behavior of the function. At these critical points, the slope of the tangent line is zero, which typically corresponds to a local maximum or minimum, or a saddle point.

Intervals of Increase and Decrease

The intervals of increase and decrease tell us where a function is rising or falling. Based on the critical points found earlier, we divide the x-axis into intervals:

• \((-\infty, -\sqrt{8})\)
• \((-\sqrt{8}, 0)\)
• \((0, \sqrt{8})\)
• \((\sqrt{8}, \infty)\)

We test each interval using a point within it to examine the sign of \(f'(x)\). If \(f'(x) > 0\), the function is increasing in that interval. If \(f'(x) < 0\), the function is decreasing. After testing, we find \(f(x)\) is increasing on \((-\infty, -\sqrt{8})\) and \((\sqrt{8}, \infty)\), and decreasing on \((-\sqrt{8}, 0)\) and \((0, \sqrt{8})\). This analysis helps visualize the overall shape and direction changes in the graph.

Polynomial End Behavior

Understanding the end behavior of a polynomial function is crucial for sketching its graph. The end behavior describes how the function behaves as \(x\) approaches positive or negative infinity. It is mainly determined by the function's leading term, which in this case is \(x^4\). For \(f(x) = x^4 - 16x^2\), as \(x\rightarrow \infty\) or \(x \rightarrow -\infty\), the \(x^4\) term dominates, causing the function to rise to \( \infty\) regardless of direction. This symmetric end behavior is characteristic of even-degree polynomials with positive leading coefficients. The graph will approach infinity on both sides, showing how polynomial end behavior provides a framework within which the details of increasing and decreasing intervals fit.