Problem 9 - Calculus with Concepts in Calculus

Question

Let \(f(x)=\frac{x-1}{x^{2}+1}\) and \(g(x)=x^{1 / 4}\). Find the specified values. $$ (f+g)(16) $$

Step-by-Step Solution

Answer

\((f+g)(16) = \frac{529}{257}\)

Step 1: Understand the Problem

We need to find the value of \((f+g)(16)\). This means we need to evaluate the sum of the functions \(f(x)\) and \(g(x)\) at \(x = 16\).

Step 2: Compute \(f(16)\)

To find \(f(16)\), substitute \(x = 16\) into the function \(f(x) = \frac{x-1}{x^2+1}\):\[ f(16) = \frac{16-1}{16^2+1} = \frac{15}{257} \]

Step 3: Compute \(g(16)\)

Next, calculate \(g(16)\) by substituting \(x = 16\) into the function \(g(x) = x^{1/4}\):\[ g(16) = 16^{1/4} = (2^4)^{1/4} = 2 \]

Step 4: Add \(f(16)\) and \(g(16)\)

Now, add the results from Steps 2 and 3 to find \((f+g)(16)\):\[ (f+g)(16) = f(16) + g(16) = \frac{15}{257} + 2 \]

Step 5: Simplify the Sum

Convert 2 into a fraction with the common denominator 257 to add:\[ 2 = \frac{514}{257} \]Then, add the two fractions:\[ (f+g)(16) = \frac{15}{257} + \frac{514}{257} = \frac{529}{257} \]

Step 6: Final Result

The expression \((f+g)(16)\) simplifies to:\[ (f+g)(16) = \frac{529}{257} \]

Step 7: Verify Simplification

Since both 529 and 257 are divisible by 1 only, \(\frac{529}{257}\) cannot be simplified further. Thus, it is the final answer.

Key Concepts

Function Evaluation

When exploring functions, function evaluation is a key concept. It involves replacing the variable in a function's expression with a given number or expression to calculate its output. For instance, in this problem, we needed to evaluate two functions, \( f(x) = \frac{x-1}{x^2+1} \) and \( g(x) = x^{1/4} \), at a specific point.

• To evaluate \( f(x) \) at \( x = 16 \), we substituted 16 into the function and calculated its result. We obtained \( f(16) = \frac{15}{257} \).
• Similarly, evaluating \( g(x) \) at the same point involves substituting 16, resulting in \( g(16) = 2 \).

The process of function evaluation is essential for understanding how functions behave at particular values, helping to reach solutions for composite functions as we see in this exercise.

Rational Functions

A rational function is a type of function that is expressed as the quotient of two polynomials. In this exercise, we dealt with the rational function \( f(x) = \frac{x-1}{x^2+1} \). Rational functions can exhibit interesting behaviors and characteristics:

• **Domain:** It includes all real numbers except those that make the denominator zero. For \( f(x) \), no real number makes \( x^2+1=0 \), so all real numbers are in its domain.
• **Simplification:** Often, rationals need simplification to make calculations easier. Though \( f(16) \) resulted directly in \( \frac{15}{257} \), other cases might require factorization or reducing terms.
• **Behavior:** Understanding rational functions' graphs can be complex with asymptotes and zeros, which is why function evaluation as in Step 2, is crucial for precise function values.

Understanding rational functions involves diving deeper into these nuances, helping in comprehending their real-world applications like rates and ratios.

Exponents and Roots

Exponents and roots provide a compact way to express large numbers and inverse relationships. In this exercise, the function \( g(x) = x^{1/4} \) involves using roots:

• **Exponents:** Represent repeated multiplication. For example, \( 2^4 = 16 \) shows multiplying 2 by itself four times.
• **Roots:** The inverse operation. The expression \( x^{1/4} \) tells us to take the fourth root of \( x \). Consequently, \( g(16) = 16^{1/4} = 2 \), since 2 multiplied by itself four times gives 16.
• **Simplification:** Recognizing that \( (2^4)^{1/4} \) simplifies directly to 2 simplifies calculation processes significantly.

These principles of exponents and roots aid in not just function operations but also in simplifying complex algebraic expressions, essential for higher-level mathematics. By unlocking these fundamentals, students can smoothly transition to more advanced topics.