Problem 8 - Applied Calculus

Question

Determine whether or not the function is a power function. If it is a power function, write it in the form \(y=k x^{p}\) and give the values of \(k\) and \(p\) $$y=\frac{5}{2 \sqrt{x}}$$

Step-by-Step Solution

Answer

Yes, it's a power function: \( y = \frac{5}{2}x^{-1/2} \), where \( k = \frac{5}{2} \) and \( p = -\frac{1}{2} \).

Step 1: Identify the Form of the Function

The first step is to look at the given function \( y = \frac{5}{2 \sqrt{x}} \) and identify its current form. "Square root" implies an exponent of \( \frac{1}{2} \), so \( \sqrt{x} = x^{1/2} \). Thus, the denominator is \( 2x^{1/2} \).

Step 2: Simplify the Expression

Rewrite the function as \( y = \frac{5}{2} \cdot \frac{1}{x^{1/2}} \). According to the laws of exponents, \( \frac{1}{x^{1/2}} = x^{-1/2} \). Therefore, the function simplifies to \( y = \frac{5}{2}x^{-1/2} \).

Step 3: Determine if the Function is a Power Function

A power function can be written as \( y = kx^p \). The simplified expression \( y = \frac{5}{2}x^{-1/2} \) fits this form where \( k = \frac{5}{2} \) and \( p = -\frac{1}{2} \).

Step 4: Verify the Result

Now that the function is written in the form of a power function, verify again: \( y = \frac{5}{2}x^{-1/2} \) matches \( y = kx^p \) with \( k = \frac{5}{2} \) and \( p = -\frac{1}{2} \). This confirms it is indeed a power function.

Key Concepts

Laws of exponents

When working with exponents, understanding their laws is crucial. These laws help us manipulate and simplify expressions involving powers. Here are some basic laws:

• Product of Powers: When multiplying two expressions with the same base, add their exponents. For instance, \( x^a \cdot x^b = x^{a+b} \).
• Quotient of Powers: When dividing two expressions with the same base, subtract their exponents: \( \frac{x^a}{x^b} = x^{a-b} \).
• Power of a Power: When raising an exponent to another exponent, multiply the exponents: \( (x^a)^b = x^{a\cdot b} \).
• Negative Exponents: A negative exponent indicates the reciprocal of the base raised to the opposite positive exponent: \( x^{-a} = \frac{1}{x^a} \).

Applying these laws allows us to rewrite expressions in simpler forms. For instance, in our given function \( y = \frac{5}{2} \cdot \frac{1}{x^{1/2}} \), applying the negative exponent law, translates it into \( y = \frac{5}{2} x^{-1/2} \). This makes the expression easier to handle and understand.

Function simplification

Function simplification involves reducing complex mathematical expressions into simpler forms. This simplifies calculation and interpretation.

• Factor Out Constants: Identify constant factors, and separate them from variable expressions. In the function \( y = \frac{5}{2} \cdot \frac{1}{x^{1/2}} \), the constant \( \frac{5}{2} \) is factored out.
• Apply Exponent Laws: Use laws like the negative exponent rule to further simplify. \( \frac{1}{x^{1/2}} \) becomes \( x^{-1/2} \).
• Combine Terms: If possible, merge like terms to simplify further. In this case, there are no like terms to combine.

After simplification, the function \( y = \frac{5}{2} x^{-1/2} \) is easier to work with. It perfectly represents the original function in a straightforward manner. Simplification is not just a mechanical process but also helps in better understanding the structure and behavior of a function.

Square root as exponent

The square root is one of the most common roots and it can be expressed using exponents. Understanding this can make dealing with radicals much easier.

• Square Root to Exponent: A square root is equivalent to an exponent of \( \frac{1}{2} \). Therefore, \( \sqrt{x} = x^{1/2} \). This representation is useful in algebraic manipulations.
• Applying to Functions: In the function \( y = \frac{5}{2 \sqrt{x}} \), the denominator \( \sqrt{x} \) complicates the expression. Rewriting it as \( x^{1/2} \) simplifies the structure and facilitates further transformation.
• Further Simplification: Using additional laws of exponents, like \( \frac{1}{x^{1/2}} = x^{-1/2} \), further simplifies and reveals the essence of the expression.

By converting roots to exponents, it becomes much easier to apply exponent laws and simplify functions. This is particularly helpful when you want to convert a function to the standard power function form, \( y = k x^p \).