Problem 23 - 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 \(f\) $$f(x)=x^{3}+2 x^{2}-4 x-8$$

Step-by-Step Solution

Answer

\(f(x) > 0\) on \((2, \infty)\); \(f(x) < 0\) on \((-\infty, -2) \cup (-2, 2)\).

Step 1: Factor the Polynomial

\(f(x) = x^3 + 2x^2 - 4x - 8\)
Group: \(= x^2(x + 2) - 4(x + 2) = (x^2 - 4)(x + 2) = (x-2)(x+2)(x+2) = (x-2)(x+2)^2\)

Step 2: Find the Zeros

\(x - 2 = 0 \Rightarrow x = 2\)
\((x + 2)^2 = 0 \Rightarrow x = -2\)

Step 3: Determine Sign Intervals

Since \((x+2)^2 \geq 0\) for all \(x\), the sign of \(f(x)\) depends on \((x - 2)\):
For \(x < -2\): \((x-2) < 0\) and \((x+2)^2 > 0\), so \(f(x) < 0\).
At \(x = -2\): \(f(-2) = 0\).
For \(-2 < x < 2\): \((x-2) < 0\) and \((x+2)^2 > 0\), so \(f(x) < 0\).
At \(x = 2\): \(f(2) = 0\).
For \(x > 2\): \((x-2) > 0\) and \((x+2)^2 > 0\), so \(f(x) > 0\).

Step 4: State the Answer

\(f(x) > 0\) for \(x \in (2, \infty)\).
\(f(x) < 0\) for \(x \in (-\infty, -2) \cup (-2, 2)\).

Key Concepts

Derivative

In calculus, the derivative of a function tells us how the function's value changes as the input changes. For a polynomial function like our given example, the derivative helps identify points where the function's slope is zero, which is crucial for finding critical points. The derivative of the function \( f(x) = x^3 + 2x^2 - 4x - 8 \) is found by applying the power rule. This rule states that for any term \( ax^n \), the derivative is \( n \cdot ax^{n-1} \). Thus, the derivative of our function is \( f'(x) = 3x^2 + 4x - 4 \). This tells us how steep the curve is at any point \(x\), and finding where \( f'(x) = 0 \) will give us the critical points where the curve might change direction.

Critical Points

Critical points are values of \(x\) where the derivative of the function is zero or undefined. These points are important because they can indicate potential maxima or minima of the function or points at which the curve changes direction. To find the critical points of our polynomial \( f(x) = x^3 + 2x^2 - 4x - 8 \), we solve \( f'(x) = 0 \):

• Consider the derivative \( f'(x) = 3x^2 + 4x - 4 \).
• Set the derivative equal to zero: \( 3x^2 + 4x - 4 = 0 \).
• Solve for \(x\) using the quadratic formula.

The solutions to this equation, \( x = \frac{2}{3} \) and \( x = -2 \), are the critical points of the function. These are where the slope of the tangent to the curve is zero, implying potential peaks or troughs on the curve.

Quadratic Formula

When finding roots of quadratic equations, we often use the quadratic formula. This formula is applicable to equations in the form \( ax^2 + bx + c = 0 \). It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our function’s derivative, \( f'(x) = 3x^2 + 4x - 4 \), we apply the quadratic formula where \( a = 3 \), \( b = 4 \), and \( c = -4 \). By substituting these values, we find:

• Calculate the discriminant: \( b^2 - 4ac = 16 + 48 = 64 \).
• Solve for \(x\): \( x = \frac{-4 \pm \sqrt{64}}{6} \).

The solutions \( x = \frac{2}{3} \) and \( x = -2 \) are the roots where the slope of our function is zero. This is crucial for determining changes in the curvature or identifying extrema on the graph of the function.

Curve Sketching

Curve sketching involves plotting the graph of a function to understand its general shape and important features. For the polynomial \( f(x) = x^3 + 2x^2 - 4x - 8 \), we use information from derivatives and critical points to assist in graphing.

• Identify where \( f(x) \) is positive or negative by evaluating the function at intervals between critical points.
• Use critical points \( x = \frac{2}{3} \) and \( x = -2 \) to locate potential maxima or minima.

Examining the signs of the derivative around these critical points, known as the first derivative test, tells us where the function is increasing or decreasing. Points of inflection, where the curve changes concavity, can provide extra insight but require the second derivative. Once these key features are identified, sketching the graph becomes a matter of plotting these behaviors accurately to depict the polynomial's curve.