Problem 1 - 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 \(p\). $$y=5 \sqrt{x}$$

Step-by-Step Solution

Answer

Yes, it is a power function with \(k = 5\) and \(p = \frac{1}{2}\).

Step 1: Write the Function in an Equivalent Algebraic Form

The given function is \(y = 5\sqrt{x}\). The square root can be expressed in exponent form as \(x^{1/2}\). Therefore, the function can be rewritten as \(y = 5x^{1/2}\).

Step 2: Compare with the Power Function Form

The standard form of a power function is \(y = kx^p\), where \(k\) and \(p\) are constants. Comparing \(y = 5x^{1/2}\) with \(y = kx^p\), we identify \(k = 5\) and \(p = \frac{1}{2}\).

Step 3: Conclusion

Since the function can be expressed in the form \(y = kx^p\) with specific values for \(k\) and \(p\), the function \(y = 5\sqrt{x}\) is indeed a power function.

Key Concepts

Exponentiation

Exponentiation is a mathematical operation that involves raising a number to the power of another number. In simpler terms, it is repeated multiplication of a base by itself an indicated number of times. The general form of exponentiation is expressed as \( a^b \), where \( a \) is the base and \( b \) is the exponent or power.
For example, \( 3^2 = 3 \times 3 = 9 \). Here, 3 is the base, and 2 is the exponent, indicating that 3 should be multiplied by itself once.

• An exponent of 1 leaves the base unchanged: \( a^1 = a \).
• An exponent of 0 gives 1 as a result except for when the base is 0: \( a^0 = 1 \), provided \( a eq 0 \).

In the context of fractional exponents, like in our problem, an exponent of \( \frac{1}{2} \) corresponds to the square root. For instance, \( x^{1/2} \) is the same as \( \sqrt{x} \). By knowing this, it's easier to transform functions involving roots to an equivalent exponential form, which is often more useful for algebraic manipulation.

Algebraic Manipulation

Algebraic manipulation involves rewriting expressions in a different but equivalent form to simplify, solve, or compare them as needed. Key processes in algebraic manipulation include factoring, expanding, and rearranging terms.

• Factoring involves breaking down an expression into a product of simpler factors. For example, rewriting \( x^2 + 2x + 1 \) as \( (x+1)^2 \).
• Expanding involves distributing and simplifying expressions like \( (x+2)^2 = x^2 + 4x + 4 \).
• Rearranging terms often helps to match the form needed for specific applications, like converting \( 5\sqrt{x} \) into the power function form \( 5x^{1/2} \).

In our example, the main algebraic manipulation is rewriting \( \sqrt{x} \) as \( x^{1/2} \). This transformation allows us to identify the function as a power function and see the constants \( k = 5 \) and \( p = 1/2 \).

Mathematical Functions

A mathematical function represents a relationship between a set of inputs and a set of permissible outputs, defined by a specific rule or expression. In simplest terms, it's like a machine that takes an input and gives out an output according to the defined rule. Functions can be linear, quadratic, exponential, or power functions, among others.
Power functions are a specific type of mathematical function, expressed in the form \( y = kx^p \) where \( k \) and \( p \) are constants. These functions describe scenarios where one quantity varies as a constant power of another. An everyday example of power functions is the area of a square \( A = s^2 \), where the area is a function of the side length squared.

• If \( p = 1 \), the function is linear, representing a direct relationship between variables.
• If \( p = 2 \), the function is quadratic, commonly seen in physics as motion equations.
• Fractional powers, like \( p = 1/2 \), indicate roots, such as the square root.

For the function \( y = 5\sqrt{x} \), rewriting it as \( y = 5x^{1/2} \) showcases it as a power function since it fits the form \( y = kx^p \), with \( k = 5 \) and \( p = 1/2 \). Recognizing functions in this standard form is crucial for identifying relationships and solving problems in algebra and calculus.