Problem 26 - 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 \(f\) $$f(x)=-x^{4}+12 x^{2}-27$$

Step-by-Step Solution

Answer

\(f(x) > 0\) for \(-\sqrt{6} < x < 0\) or \(0 < x < \sqrt{6}\); \(f(x) < 0\) for \(x < -\sqrt{6}\) or \(x > \sqrt{6}\).

Step 1: Identify Critical Points

First, we need to find the critical points of the function where the derivative is zero or undefined. Calculate the derivative by using power rule: \( f'(x) = -4x^3 + 24x \).Set the derivative equal to zero and solve for \(x\):\( -4x^3 + 24x = 0 \).Factor the equation: \( -4x(x^2 - 6) = 0 \).The critical points are \( x = 0 \) and \( x = \pm \sqrt{6} \).

Step 2: Analyze Intervals of Critical Points

We will analyze the intervals defined by the critical points: \((-\infty, -\sqrt{6})\), \((-\sqrt{6}, 0)\), \((0, \sqrt{6})\), and \((\sqrt{6}, \infty)\).For each interval, select a test point, plug it into the derivative \(f'(x) = -4x^3 + 24x\), and determine the sign:- Interval \((-\infty, -\sqrt{6})\): Test point \( x = -3 \) gives \( f'(-3) < 0 \) (function decreasing).- Interval \((-\sqrt{6}, 0)\): Test point \( x = -1 \) gives \( f'(-1) > 0 \) (function increasing).- Interval \((0, \sqrt{6})\): Test point \( x = 1 \) gives \( f'(1) > 0 \) (function increasing).- Interval \((\sqrt{6}, \infty)\): Test point \( x = 3 \) gives \( f'(3) < 0 \) (function decreasing).

Step 3: Determine Sign of Function in Each Interval

Evaluate \(f(x)\) at critical points and be sure it's positive or negative in each interval:- Interval \((-\infty, -\sqrt{6})\): Choose \( x = -3 \), \( f(-3) = -81 + 108 - 27 = 0 \).- Interval \((-\sqrt{6}, 0)\): Choose \( x = -2 \), \( f(-2) = -16 + 48 - 27 = 5 \) (function positive).- Interval \((0, \sqrt{6})\): Choose \( x = 2 \), \( f(2) = -16 + 48 - 27 = 5 \) (function positive).- Interval \((\sqrt{6}, \infty)\): Choose \( x = 3 \), \( f(3) = -81 + 108 - 27 = 0 \). The signs of the function are analyzed around the intervals to state when \( f(x) > 0 \).

Step 4: Solution

The function \( f(x) = -x^4 + 12x^2 - 27 \) is positive when \( -\sqrt{6} < x < 0 \) or \( 0 < x < \sqrt{6} \). It's negative when \( x < -\sqrt{6} \) or \( x > \sqrt{6} \).

Step 5: Graph Sketching

Plot the critical points \( x = 0 \) and \( x = \pm \sqrt{6} \), and analyze the behavior by checking the changes in signs:- The graph slopes down in \((-\infty, -\sqrt{6})\), goes up from \((-\sqrt{6}, 0)\), continues up from \((0, \sqrt{6})\), and slopes down \((\sqrt{6}, \infty)\).- Function values at critical points: \( f(0) = -27 \), \( f(\pm \sqrt{6}) = 9 \).These values mark turning points as maxima or minima on the graph.

Key Concepts

Critical Points

Critical points are crucial in understanding the behavior of a function. These points are where the function's derivative is zero or undefined. It indicates a potential change in how the function is increasing or decreasing.

To locate the critical points of the function \( f(x) = -x^4 + 12x^2 - 27 \), we first compute the derivative. Using the power rule, we find:

• \( f'(x) = -4x^3 + 24x \)

Setting this derivative equal to zero, we solve for \( x \):

\( -4x(x^2 - 6) = 0 \)

This gives us the critical points at \( x = 0 \) and \( x = \pm \sqrt{6} \). These points divide the function into different regions, each needing further analysis to see what happens at these boundaries.

Derivative

The derivative of a function is a powerful tool in calculus. It measures how a function changes as its input changes, giving the rate of change or slope at any point.

For our function \( f(x) = -x^4 + 12x^2 - 27 \), the derivative is \( f'(x) = -4x^3 + 24x \). This derivative helps us determine how the function is behaving between and at the critical points. By examining the sign of the derivative, we can tell if the function is increasing or decreasing over certain intervals:

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

Testing points within intervals defined by the critical points helps validate the function behavior. This step is crucial for sketching the graph accurately.

Intervals

Interval analysis involves breaking the domain of the function into sections determined by the critical points. For each interval, we determine whether the function is positive or negative.

The critical points \( x = 0 \) and \( x = \pm \sqrt{6} \) divide the real number line into four intervals:

• \((-\infty, -\sqrt{6})\)
• \((-\sqrt{6}, 0)\)
• \((0, \sqrt{6})\)
• \((\sqrt{6}, \infty)\)

In each interval, we choose a test point and evaluate \( f(x) \) and \( f'(x) \):

• In \((-\infty, -\sqrt{6})\), selecting \( x = -3 \) shows \( f'(x) < 0 \), and \( f(x) = 0 \), the function is non-positive.
• In \((-\sqrt{6}, 0)\) and \((0, \sqrt{6})\), test points show \( f(x) > 0 \) and \( f'(x) > 0 \).
• In \((\sqrt{6}, \infty)\), \( f(x) = 0 \) and \( f'(x) < 0 \), indicating a return to non-positive values.

Understanding how the function behaves in these intervals tells us where it is positive or negative.

Graph Sketching

Graph sketching is the visual representation of a function's behavior over its domain. It integrates the information from the critical points, derivative analysis, and interval sign checks.

First, plot the critical points: \( x = 0 \) and \( x = \pm \sqrt{6} \). The value at these points and their corresponding function values \( f(0) = -27 \), \( f(\pm \sqrt{6}) = 9 \) tell you where potential maxima or minima occur:

• \( x = 0 \) indicates a minimum since \( f(x) \) transitions from increasing to decreasing.
• \( x = \pm \sqrt{6} \) are maxima with positive \( f(x) \) values in the surrounding intervals.

By analyzing changes in the derivative's sign, draw slopes that reflect increasing or decreasing trends between these critical points. This results in the graph given by:

• Downward slope as \( x \to -\infty \), peaking upwards to \( x = -\sqrt{6} \).
• Rising towards zero, then again towards \( x = \sqrt{6} \).
• Finally, the graph dips after \( x = \sqrt{6} \) as \( x \to \infty \).

The completed sketch now reflects these changes, with maxima and minima appropriately marked.