Problem 18 - 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 \(f\) $$f(x)=9 x-x^{3}$$

Step-by-Step Solution

Answer

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

Step 1: Find the Critical Points

To find the critical points of the function, we first take the derivative of \(f(x) = 9x - x^3\). The derivative, \(f'(x)\), is found as follows: \[ f'(x) = \frac{d}{dx}(9x - x^3) = 9 - 3x^2. \]Set \(f'(x) = 0\) to find the critical points:\[ 9 - 3x^2 = 0 \]Solving for \(x\) gives:\[ 3x^2 = 9 \]\[ x^2 = 3 \]\[ x = \pm \sqrt{3}. \] Thus, the critical points are \(x = \sqrt{3}\) and \(x = -\sqrt{3}\).

Step 2: Determine Intervals and Test Points

We use the critical points to divide the number line into intervals: \((-\infty, -\sqrt{3}), (-\sqrt{3}, \sqrt{3}), (\sqrt{3}, \infty)\).Choose test points from each interval to determine where \(f(x)\) is positive and negative.- For \((-\infty, -\sqrt{3})\), pick \(x = -2\).- For \((-\sqrt{3}, \sqrt{3})\), pick \(x = 0\).- For \((\sqrt{3}, \infty)\), pick \(x = 2\).

Step 3: Evaluate Test Points

Evaluate \(f(x)\) at each test point:- For \(x = -2\): \[ f(-2) = 9(-2) - (-2)^3 = -18 + 8 = -10 < 0. \]- For \(x = 0\): \[ f(0) = 9(0) - 0^3 = 0. \]- For \(x = 2\): \[ f(2) = 9(2) - 2^3 = 18 - 8 = 10 > 0. \]

Step 4: Determine Sign of f(x) on Each Interval

Using the evaluations:- On \((-\infty, -\sqrt{3})\), \(f(x) < 0\).- On \((-\sqrt{3}, \sqrt{3})\), \(f(x) = 0\).- On \((\sqrt{3}, \infty)\), \(f(x) > 0\).

Step 5: Sketch the Graph and Summarize Values

To sketch the graph, plot the points where \(f(x) = 0\) (i.e., \((-3, 0)\) and \((3, 0)\)) and the critical points \(x = \pm \sqrt{3}\) with their respective function values. \[f(x) > 0\] on \((\sqrt{3}, \infty)\) and \[f(x) < 0\] on \((-\infty, -\sqrt{3})\).The graph opens down like a cubic, passing through the above points.

Key Concepts

Derivative of a function

The derivative of a function gives us crucial insights about the behavior of that function. In essence, the derivative tells us how the function's output values change in response to changes in the input values. For a given function like \( f(x) = 9x - x^3 \), the derivative represented as \( f'(x) \), provides information about the slope or rate of change at any point along its curve.
The derivative can be found using well-defined differentiation rules. For the function \( f(x) = 9x - x^3 \), we apply the power rule to find that:

• \( f'(x) = 9 - 3x^2 \)

Setting the derivative equal to zero, \( 9 - 3x^2 = 0 \), allows us to solve for the critical points. These are the values of \( x \) where the function's slope is either at a maximum, minimum, or horizontal inflection point, specifically, \( x = \pm \sqrt{3} \). These critical points divide the function into distinct intervals where the behavior of \( f(x) \) can be analyzed further.
Understanding derivatives is an essential part of calculus and helps with tasks like finding maximum and minimum values, optimizing properties, and understanding the geometry of graphs.

Intervals of increase and decrease

To understand where a function increases or decreases, we use the critical points derived from the function's derivative. These points help segment the x-axis into intervals where the function's behavior can be evaluated easily. For example, with critical points \( x = \pm \sqrt{3} \), the function \( f(x) = 9x - x^3 \) is divided into these intervals: \((-\infty, -\sqrt{3})\), \((-\sqrt{3}, \sqrt{3})\), and \((\sqrt{3}, \infty)\).
Choosing test points within these intervals lets us determine the function's nature in each segment. Testing these regions with points like \(-2, 0,\) and \(2\):

• For \( x = -2 \) (in \((-\infty, -\sqrt{3})\)), \( f(x) < 0 \), implying the function is decreasing.
• For \( x = 0 \) (in \((-\sqrt{3}, \sqrt{3})\)), \( f(x) = 0 \), displaying a neutral or turning behavior at this critical area.
• For \( x = 2 \) (in \((\sqrt{3}, \infty)\)), \( f(x) > 0 \), indicating the function is increasing.

Maps of increasing and decreasing intervals of a function aid in thoroughly graphing and understanding how the function behaves across various sections of the input range. Recognizing these intervals is also critical for solving optimization problems and comprehending the full scope of a function's graph.

Sketching graphs of functions

Skilled graph sketching involves understanding a function's behavior as depicted by its critical points, derivatives, and intervals of increase or decrease. For our function \( f(x) = 9x - x^3 \), the process begins with graphing critical points such as \( x = \pm\sqrt{3} \). These provide a framework by representing points where the slope changes or the function takes a notable turn.
Next, we evaluate where the function crosses the x-axis, known as the roots. Points like \((-3, 0)\) and \((3, 0)\) are vital as they mark intersections with the x-axis. Combined with the previously identified intervals:

• The graph is below the x-axis from \((-\infty, -\sqrt{3})\)
• The graph crosses or touches the x-axis at \((-\sqrt{3}, \sqrt{3})\)
• The graph is above the x-axis from \((\sqrt{3}, \infty)\)

This analysis results in a refined sketch that portrays the essence of the function: it manifests like a cubic curve with two turning points, indicating where the graph "humps". Better sketches rely on accurately understanding these properties, thus allowing us to see if the function is symmetric, has any asymptotes or peculiar features, and visually conceptualize its overall trend.