Problem 36 - Precalculus Mathematics for Calculus

Question

Solve the equation both algebraically and graphically. $$x^{3}+16=0$$

Step-by-Step Solution

Answer

The solution to \(x^3 + 16 = 0\) is \(x = -2.52\).

Step 1: Understand the Problem

The problem asks us to solve the equation \(x^3 + 16 = 0\) both algebraically and graphically. This means we need to find the value(s) of \(x\) that satisfy this equation using algebraic manipulation and by visualizing it on a graph.

Step 1: Solve Algebraically

To solve \(x^3 + 16 = 0\) algebraically, start by isolating \(x^3\): \[ x^3 + 16 = 0 \] Subtract 16 from both sides: \[ x^3 = -16 \] Now, take the cube root of both sides: \[ x = \sqrt[3]{-16} \] The cube root of \(-16\) is \(-2.52\), which means \(x = -2.52\) is the solution algebraically.

Step 2: Solve Graphically

To solve the equation graphically, we need to plot the function \(f(x) = x^3 + 16\) and identify where the graph intersects the x-axis. The x-axis intersection points are the solution(s) to the equation:1. Plot \(y = x^3\), which is a cubic function with a shape similar to an S-curve.2. Shift the curve vertically up (or down) by 16 units to represent \(f(x) = x^3 + 16\).3. Observe the graph intersecting the x-axis at the point where \(x \approx -2.52\).

Step 3: Verify the Solution

To verify, substitute \(x = -2.52\) back into the original equation:\[ (-2.52)^3 + 16 = 0 \] Calculate \((-2.52)^3 = -16\); hence, \[ -16 + 16 = 0 \] This confirms that \(x = -2.52\) satisfies the equation.

Key Concepts

Cubic Functions

Cubic functions are polynomial functions of degree three. They are written in the form \( ax^3 + bx^2 + cx + d = 0 \) where \( a eq 0 \). The equation given in the problem, \( x^3 + 16 = 0 \), is a cubic function with no quadratic or linear terms. This simplifies our problem as it focuses on the cubic term and the constant. Characteristics of cubic functions:

• The graph of a cubic function is an S-shaped curve, which may have one real root and up to two complex roots.
• The end behavior of the polynomial depends on the leading coefficient. For positive leading coefficients, the graph extends from negative to positive infinity; for negative, it extends from positive to negative infinity.
• Cubic functions can have at most one inflection point.

Understanding these fundamental properties helps in visualizing and manipulating cubic functions both algebraically and graphically.

Algebraic Solutions

To find algebraic solutions to polynomial equations like cubic functions, we use various algebraic techniques to isolate the variable or manipulate the equation into a simpler form.In this exercise, we isolated \( x \) in the equation \( x^3 + 16 = 0 \) by first subtracting 16 from both sides. This yields:\[ x^3 = -16 \]Then, to solve for \( x \), we take the cube root of both sides. Cube roots can easily be calculated with a calculator or algebraic manipulation:\[ x = \sqrt[3]{-16} \approx -2.52 \]Solving cubic equations can sometimes yield multiple solutions, especially when involving complex numbers. However, since this cubic equation results in a real-valued root, our focus remains on the real number solution. This simple approach simplifies the task of finding solutions to cubic functions.

Graphical Solutions

Graphical solutions to polynomial equations involve plotting the function and observing where it intersects the x-axis. Each intersection represents a solution to the equation.For the exercise \( x^3 + 16 = 0 \), we plotted:

• The function \( f(x) = x^3 + 16 \)
• Notice the upward S-curve of \( y = x^3 \)

Then, shift this graph vertically downward by 16 units to simulate the adjustment made in \( \log(y = x^3 + 16) \). The next step was to visually identify the x-intercepts, where the function crosses the x-axis.Here, the graph of \( f(x) = x^3 + 16 \) intersects the x-axis at approximately \( x = -2.52 \). This graphical approach makes it easier to visualize solutions, especially for confirming results obtained algebraically. It's important because students can see the behavior of polynomial functions and precisely where they hit the x-axis.