Problem 14 - Calculus with Concepts in Calculus

Question

Find the numerical value of the function at the given values of \(a\). $$ g(x)=4.5 x^{1 / 2}-x^{3 / 2} ; a=3,1.64 $$

Step-by-Step Solution

Answer

\( g(3) \approx 2.598 \) and \( g(1.64) \approx 3.14 \).

Step 1: Understand the Function

The given function is \( g(x) = 4.5 x^{1/2} - x^{3/2} \). It combines two expressions: one involving \( x^{1/2} \) (which is the square root of \( x \)) and another involving \( x^{3/2} \) (which is \( x \) raised to the power of 3/2). Our task is to evaluate this function at specific values of \( a \).

Step 2: Evaluate the Function at a = 3

First, substitute \( a = 3 \) into the function: \( g(3) = 4.5(3)^{1/2} - (3)^{3/2} \). Calculate each term separately. \( 3^{1/2} = \sqrt{3} \approx 1.732 \) and \( 3^{3/2} = (\sqrt{3})^3 \approx 5.196 \). Substituting these estimates into the function gives: \( g(3) \approx 4.5 \times 1.732 - 5.196 \approx 7.794 - 5.196 \approx 2.598 \).

Step 3: Evaluate the Function at a = 1.64

Next, substitute \( a = 1.64 \) into the function: \( g(1.64) = 4.5(1.64)^{1/2} - (1.64)^{3/2} \). Start by calculating \( 1.64^{1/2} = \sqrt{1.64} \approx 1.28 \) and \( 1.64^{3/2} = (\sqrt{1.64})^3 \approx 2.62 \). Substituting these results into the function gives: \( g(1.64) \approx 4.5 \times 1.28 - 2.62 \approx 5.76 - 2.62 \approx 3.14 \).

Key Concepts

Function Evaluation

Function evaluation is like playing detective with math. It involves finding out what a function gives us for specific input values. In our exercise, the function is given as \[ g(x) = 4.5 x^{1/2} - x^{3/2} \]This means for any input \( x \), you substitute \( x \) into the function and simplify it to find the output. It's similar to putting a number into a vending machine and seeing what item comes out.
For example, when evaluating this function for \( a = 3 \), we replace \( x \) with 3 and then calculate the result:

• First term: \( 4.5 \times (3)^{1/2} \)
• Second term: \( -(3)^{3/2} \)

Then, we subtract the second term from the first to get the result. This is essentially what we mean by function evaluation!

Exponents

Exponents are a way of writing repeated multiplication. When you see something like \( x^{n} \), it's telling you to multiply \( x \) by itself \( n \) times. In our function, we see exponents such as \( x^{1/2} \) and \( x^{3/2} \).
Here's what these mean:

• \( x^{1/2} \): This is the square root of \( x \).
• \( x^{3/2} \): This is first taking the square root of \( x \) (i.e., \( x^{1/2} \)) and then cubing the result (i.e., raising it to the power of 3).

Understanding these exponents is crucial because they transform our input \( x \) in the specified ways and influence the function's output. They show us how compactly we can represent complex operations.

Square Root

The square root is a special exponent - it's like asking "what number, when multiplied by itself, gives us this number?" So, \( \sqrt{x} = x^{1/2} \).
Square roots appear frequently in our function, specifically in the component \( x^{1/2} \). This part of the function asks us to determine what number, multiplied by itself, sums to \( x \).
For example, for \( x = 3 \), the square root \( \sqrt{3} \approx 1.732 \). It's crucial to know how to find square roots since our function relies on accurately calculating them for evaluating the expression at any given \( a \).
Using the square root properly helps us break down problems into smaller, more manageable parts and is an essential skill in calculus.