Problem 13 - 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. $$ f(g(1)) $$

Step-by-Step Solution

Answer

The value of \( f(g(1)) \) is 0.

Step 1: Evaluate the inner function

First, we need to evaluate the inner function, which is \( g(1) \). Since \( g(x) = x^{1/4} \), substitute \( x = 1 \) into the function. We have:\[g(1) = 1^{1/4} = 1\]Thus, \( g(1) = 1 \).

Step 2: Substitute the result into the outer function

Now that we have \( g(1) = 1 \), substitute this result into the function \( f(x) \). So we need to evaluate \( f(1) \):\[f(x) = \frac{x-1}{x^2+1}\]Substituting \( x = 1 \) gives:\[f(1) = \frac{1-1}{1^2+1} = \frac{0}{2} = 0\]Thus, \( f(g(1)) = f(1) = 0 \).

Key Concepts

Evaluating Functions

In mathematics, evaluating functions is a core concept that involves finding the output value of a function for a specific input. This operation is crucial when dealing with composite functions, where you apply one function to the results of another. In the exercise, we are asked to find the value of the composite function \( f(g(x)) \). Here is a breakdown of how this evaluation process works:

• Identify the inner function, in this case, \( g(x) \), and evaluate it using the given input: \( g(1) \).
• Calculate \( g(1) \) by substituting \( x = 1 \) into \( g(x) = x^{1/4} \).
• Once you get \( g(1) = 1 \), move to the outer function \( f(x) \).
• Substitute the result from the inner function \( g(1) \) into the outer function: \( f(1) \).
• Compute \( f(1) = \frac{1-1}{1^2+1} = 0 \), which gives us the final result, \( f(g(1)) = 0 \).

By following these steps systematically, evaluating composite functions becomes straightforward. The key is resolving the inner function first and using that outcome for the outer function.

Exponential Functions

Exponential functions are a special class of functions where the variable is in the exponent. In our example, this is seen in \( g(x) = x^{1/4} \). Here are some essential points:

• In general, exponential functions have the form \( g(x) = a^{x} \) or the inversion as seen in our exercise.
• The base, in this case, is \( x \), and the exponent is \( \frac{1}{4} \), a fractional power representing the fourth root of \( x \).
• Calculating a fourth root such as \( 1^{1/4} \) is straightforward because \( 1 \) raised to any power remains \( 1 \).

Exponential functions, while extensive in their usage, often include roots and fractional exponents in problems like the one we just solved. These require careful attention to the laws of indices but generally follow straightforward arithmetic rules.

Rational Functions

Rational functions, like \( f(x) = \frac{x-1}{x^2+1} \), are ratios of polynomial functions. Understanding them is pivotal because they frequently appear in calculus and algebra.

• A rational function is expressed as \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials.
• The domain of a rational function is all real numbers except where the denominator \( q(x) = 0 \).
• In our exercise, the denominator is \( x^2 + 1 \), which never equals zero for real numbers, so \( f(x) \) is defined for all real \( x \).
• When we substitute \( x = 1 \), the numerator becomes zero, simplifying the function \( f(x) \) to zero at this point.

Rational functions can represent complex behaviors in graphs and equations, but the focus should always be on simplifying and understanding polynomials and their roots and coefficients. They require attention to their domain and simplifying fractions wherever possible.