Question

Evaluate and simplify \(g(0.6 c)\) given that $$ \begin{array}{l} f(x)=x^{-1 / 2} \\ g(v)=f\left(1-v^{2} / c^{2}\right) \end{array} $$

Short answer

Answer: \(\frac{5}{4}\)

Step-by-step solution

Substitute \(0.6c\) into \(g(v)\)

We are given \(g(v) = f\left(1-\frac{v^2}{c^2}\right)\). Let's substitute \(0.6c\) for \(v\): $$ g(0.6c) = f\left(1 - \frac{(0.6c)^2}{c^2}\right) $$

Simplify the expression inside the parentheses

Now, we need to simplify the expression inside the parentheses: $$ 1 - \frac{(0.6c)^2}{c^2} = 1 - \frac{(0.6^2)(c^2)}{c^2} $$ Since \(0.6^2 = 0.36\) and \((c^2)/(c^2) = 1\), this simplifies further to: $$ 1 - 0.36 = 0.64 $$ Thus, we have: $$ g(0.6c) = f(0.64) $$

Substitute \(0.64\) into \(f(x)\)

Now, let's substitute \(0.64\) into the function \(f(x)\): $$ f(x) = x^{-1/2} \Rightarrow f(0.64) = 0.64^{-1/2} $$

Simplify the expression

To simplify this expression, we can calculate the square root of \(0.64\) and then find its reciprocal: $$ 0.64^{-1/2} = (\sqrt{0.64})^{-1} = (0.8)^{-1} $$ The reciprocal of \(0.8\) is: $$ (0.8)^{-1} = \frac{1}{0.8} = \frac{5}{4} $$

Write the final answer

Now that we have simplified the expression, we can write the final answer: $$ g(0.6c) = \frac{5}{4} $$

Key concepts

Substitution

Function evaluation often involves substitution, a fundamental concept in algebra. When we have a function like \( g(v) \), and we need to evaluate it at a specific value, our task is to "substitute" that value into the function. In the context of the given exercise, we substitute \( v = 0.6c \) into \( g(v) = f\left(1 - \frac{v^2}{c^2}\right) \).
Substitution is straightforward but requires precision:

• Identify the variable you are substituting.
• Replace every instance of the variable with the given value.
• Ensure that every part of the function is adjusted accordingly.

For example, when substituting \( 0.6c \) for \( v \), it changes to \( g(0.6c) = f\left(1 - \frac{(0.6c)^2}{c^2}\right) \). It directly plugs the value \( 0.6c \) into any place that originally uses \( v \).This technique is vital for determining how a function behaves at specific points, a step crucial for deeper function analysis.

Simplifying Expressions

After substitution, the next critical step in function evaluation is simplifying expressions. Simplifying makes complex equations more manageable by transforming them into more understandable terms. For our exercise, once \( g(0.6c) \) was presented as \( f\left(1 - \frac{(0.6c)^2}{c^2}\right) \), it was necessary to simplify this expression.
To simplify:
- First, calculate \( (0.6c)^2 = 0.36c^2 \). - Then, divide by \( c^2 \), resulting in \( \frac{0.36c^2}{c^2} = 0.36 \). - Finally, subtract this from 1, giving \( 1 - 0.36 = 0.64 \). Why is simplification important?
- It reveals the essential structure of an expression, allowing for easier subsequent calculations.- In functions, it ensures the domain and range can be clearly understood.- Simplification eliminates unnecessary complexity, leading to clearer results with fewer errors.This process illustrates how simplification prepares expressions for further evaluation, like plugging into other functions or performing additional operations.

Reciprocal Calculations

Reciprocal calculations are an essential component when dealing with expressions raised to negative exponents. In the given problem, we reach a point where we must compute \( 0.64^{-1/2} \). This is equivalent to finding the reciprocal of the square root of 0.64.
Here's how to tackle reciprocal calculations:

• First, interpret \( x^{-1/2} \) as \( \frac{1}{\sqrt{x}} \).
• Next, find the square root of \( x \). For 0.64, \( \sqrt{0.64} = 0.8 \).
• Finally, calculate the reciprocal: \( \frac{1}{0.8} = \frac{5}{4} \).

Reciprocal calculations are helpful because:
- They convert negative exponents into fractional representation, which can simplify algebraic manipulations.- They're key in various mathematical fields, particularly when dealing with rates or inverses of quantities.- Understanding how to handle reciprocals improves comprehension of more complex mathematical operations, empowering students to approach diverse problems confidently.