Problem 14 - Contemporary Precalculus

Question

Determine whether the equation defines \(y\) as a function of \(x\) or defines \(x\) as a function of \(y\) $$5 x-4 y^{4}+64=0$$

Step-by-Step Solution

Answer

Answer: The given equation defines both y as a function of x and x as a function of y.

Step 1: Try to express y as a function of x

To express y as a function of x, our goal is to isolate y on one side of the equation. In this case, we will try to leave y alone on the left side of the equation. $$4y^4 = 5x - 64$$ $$y^4 = \frac{5x - 64}{4}$$ $$y = \pm\sqrt[4]{\frac{5x - 64}{4}}$$ Since we are able to express y as a function of x (y is only dependent on x), we can say that the given equation defines y as a function of x.

Step 2: Try to express x as a function of y

To express x as a function of y, our goal is to isolate x on one side of the equation. We will try to leave x alone on the left side of the equation. $$5x = 4y^4 - 64$$ $$x = \frac{4y^4 - 64}{5}$$ Since we were able to express x as a function of y (x is only dependent on y), we can also say that the given equation defines x as a function of y. In conclusion, the given equation defines both \(y\) as a function of \(x\) and \(x\) as a function of \(y\).

Key Concepts

Expressing Variables

In mathematics, expressing a variable means rewriting an equation so that a specific variable stands alone on one side. This process is crucial to identify functions and understand the relationship between variables.
For example, if you start with the equation \(5x - 4y^4 + 64 = 0\), your task might be to express \(y\) in terms of \(x\).
Through algebraic manipulation, you isolate \(y\) on one side of the equation. First, move all terms involving \(x\) to the other side to get \(4y^4 = 5x - 64\). Then, you further simplify this to \(y^4 = \frac{5x - 64}{4}\). Finally, you solve for \(y\), which gives \(y = \pm \sqrt[4]{\frac{5x - 64}{4}}\).
This outcome shows how \(y\) can be expressed based on \(x\), thus defining a function from \(x\) to \(y\).

• Expressing the variable helps determine which variable is independent and which is dependent.
• Each manipulation involves carefully following algebraic rules like addition, subtraction, multiplication, and division.
• Expressing \(y\) in terms of \(x\) was possible due to the reversible operations involved.

Solving Equations

Solving equations involves finding the value(s) of the variable(s) that make the equation true. This is a foundational concept in algebra and is used to understand how variables interact.
For example, in the equation \(5x - 4y^4 + 64 = 0\), you can solve for \(y\) by isolating it on one side. Similarly, you can solve for \(x\) in terms of \(y\).
Here's how you solve for \(x\): start by rearranging the formula to get \(5x = 4y^4 - 64\). Then divide both sides by 5 to isolate \(x\), which gives \(x = \frac{4y^4 - 64}{5}\).

Key considerations in solving equations include:

• Perform operations equally on both sides of the equation to maintain balance.
• Check the validity of your solutions by substituting back into the original equation.
• Be aware of multiple possible solutions, especially when dealing with higher-degree equations as in solving for \(y\).

Solving equations shows us whether a variable defines a function of another by demonstrating dependence or independence.

Dependent and Independent Variables

Variables are often classified as either dependent or independent in an equation or function.
This classification helps us understand the relationship and interactions between variables.
In the equation \(5x - 4y^4 + 64 = 0\), both \(x\) and \(y\) can act as dependent or independent variables depending on how the equation is manipulated and what is being solved for.

• The independent variable, typically located on the right side when expressing a function, is what impacts the dependent variable. For example, if \(y\) is expressed as \(y = \pm \sqrt[4]{\frac{5x - 64}{4}}\), \(x\) is the independent variable and \(y\) is dependent on it.
• When reversed, as in \(x = \frac{4y^4 - 64}{5}\), \(y\) becomes the independent variable, defining \(x\) as the dependent variable.
• Understanding which variable is independent helps clarify how changes in one variable affect the other.

Both situations were possible in this equation since it was solvable for both cases.