Solution: Compute $ f(5) = 25 - 30 + m = -5 + m $ and $ g(5) = 25 - 30 + 3m = -5 + 3m $. - Londonproperty

Understanding the Context

In algebra, $ f(5) $ refers to substituting $ x = 5 $ into the function $ f(x) $. Similarly, $ g(5) $ means evaluating $ g(x) $ at $ x = 5 $. For the given functions:

• $ f(x) = 25 - 30 + m $
  
• $ g(x) = 25 - 30 + 3m $

Substituting $ x = 5 $ yields:

$$
f(5) = 25 - 30 + m = -5 + m
$$
$$
g(5) = 25 - 30 + 3m = -5 + 3m
$$

Key Insights

This substitution helps simplify expressions, evaluate outputs for specific inputs, and explore dependencies on parameters like $ m $.

Why Evaluate at $ x = 5 $?

Evaluating functions at specific values is essential for:

• Function prediction: Determining outputs for given inputs is useful in modeling real-world scenarios.
  
• Parameter dependence: Expressions like $ -5 + m $ and $ -5 + 3m $ show how variable $ m $ influences results.
  
• Problem solving: Substituted values help verify solutions, compare functions, and solve equations.

For example, setting $ f(5) = 0 $ allows solving for $ m = 5 $, simplifying $ f(x) $, and understanding how $ f(x) $ behaves overall.

Final Thoughts

Step-by-Step Evaluation: $ f(5) $ and $ g(5) $

Step 1: Simplify the expressions

Begin with the basic arithmetic:

$$
f(5) = (25 - 30) + m = -5 + m
$$
$$
g(5) = (25 - 30) + 3m = -5 + 3m
$$

Step 2: Substitute $ x = 5 $

As shown above, replacing $ x $ with 5 yields these expressions in terms of $ m $.

Step 3: Analyze parameter impact

The parameter $ m $ acts as a variable multiplier in $ g(5) $, amplifying its effect. In contrast, $ f(5) $ depends linearly on $ m $, making both functions sensitive yet distinct in scaling.

• If $ m = 2 $:
    $ f(5) = -5 + 2 = -3 $
    $ g(5) = -5 + 3(2) = 1 $
  
• If $ m = 5 $:
    $ f(5) = -5 + 5 = 0 $
    $ g(5) = -5 + 15 = 10 $