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Chapter 2: Problem 21

Determine whether each equation defines y as a function of x. $$ x+y^{3}=8 $$

Short Answer

Expert verified
Yes, the equation does define y as a function of x because for every x-value, there is exactly one corresponding y-value.

Step by step solution

01

Express y in terms of x

In order to express y as a function of x, the first step is to manipulate the equation to solve for y, isolating the y terms on one side of the equation. Start by subtracting x from both sides of equation \(x+y^{3}=8\). This gets \(y^{3}=8-x\)
02

Apply Cubic Root to both sides

Apply a cubic root to both sides of equation \(y^{3}=8-x\) to solve for y. When taking the cubic root of both sides, remember that the cubic root function returns three possible roots. Therefore, applying the cubic root gets \(y=\sqrt[3]{8-x}\)
03

Determine if y is a function of x

Considering \(y=\sqrt[3]{8-x}\), for every x-value, there is just one corresponding y value. Hence, this determines that y is a function of x.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Relation Between Variables
In mathematics, understanding the relationship between variables is crucial. Here, we start with the given equation:
\(x + y^3 = 8\).
Identifying this relationship helps us understand how one variable depends on another.
  • x-variable (independent): Manipulating this variable directly influences the dependent variable.
  • y-variable (dependent): Its value depends on the x-variable through the equation.
To determine if 'y' is a function of 'x', we isolate 'y' and see if for each 'x', there is a single 'y' value. This ensures a clear, consistent relationship between the variables.
Solving Equations
The next step involves solving the equation by isolating the dependent variable. Starting with the equation:
\(x + y^3 = 8\),
we aim to isolate \(y^3\) on one side.
  • Subtract 'x' from both sides: This shifts 'x' away from \(y^3\), resulting in \(y^3 = 8 - x\). This is a crucial operation as it simplifies our task of solving for \(y\).
By performing algebraic operations like addition, subtraction, multiplication, or division, you can manipulate equations to find explicit expressions of variables.
Cubic Root
To find 'y' from \(y^3 = 8 - x\), we must apply cubic rooting. Cubic roots are special because they consider each number's unique characteristics. For the equation:
\(y^3 = 8 - x\),
calculating the cubic root of both sides gives us:
\(y = \sqrt[3]{8 - x}\).
  • Handling Cubic Roots: Unlike square roots, cubic roots for real numbers yield one real number, making them simpler in many cases.
  • Unique Solution: With cubic roots, each 'x' value produces exactly one 'y', contributing to clarity and influencing our problem-solving strategy.
Cubic roots allow us to solve equations succinctly, revealing the function's specific characteristics.
Function of x
In determining whether \(y\) is a function of \(x\), we check if each \(x\) corresponds to a unique \(y\). With the obtained equation:
\(y = \sqrt[3]{8 - x}\),
for each input value of \(x\), there is precisely one output value of \(y\).
  • Definition of a Function: A function relates every element "\(x\)" in the domain to a unique element "\(y\)" in the range.
  • Graphically: Each vertical line in the "function of \(x\)" graph will intersect the curve at exactly one point, satisfying the vertical line test.
Thus, \(y = \sqrt[3]{8 - x}\) is indeed a function of \(x\), because every \(x\) has a unique 'y' output.

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Most popular questions from this chapter

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I must have made a mistake in finding the composite functions \(f \circ g\) and \(g \circ f,\) because I notice that \(f \circ g\) is not the same function as \(g \circ f\)

graph both equations in the same rectangular coordinate system and find all points of intersection. Then show that these ordered pairs satisfy the equations. $$\begin{array}{r} {x^{2}+y^{2}=9} \\ {x-y=3} \end{array}$$

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Solve and check: \(\frac{x-1}{5}-\frac{x+3}{2}=1-\frac{x}{4}\) (Section \(1.2, \text { Example } 3)\)

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