91Ó°ÊÓ

Function evaluation is the process of determining the output value of a function for a specific input value of its variable. In simpler terms, it's like asking what result you get when you "plug in" a number into a function. For the functions given, like \( f(x) = 2x^2 - 3x + 1 \) and \( g(x) = \frac{x^2 + 1}{x} \), finding the value when \( x = 1 \) involves substituting 1 for every \( x \) in the equations.

• Start by replacing \( x \) with 1 in both functions.
• Calculate each term separately to avoid mistakes.
• Combine the calculated terms to find the final result for that specific input.

In our example with \( f(x) \), substitute 1 into the expression to get \( f(1) = 2(1)^2 - 3(1) + 1 = 0 \). Similarly, for \( g(x) \), substitute to find \( g(1) = \frac{1^2 + 1}{1} = 2 \). A solid understanding of function evaluation is crucial for more advanced calculus concepts as well.

Rational functions are a category of functions defined as the ratio of two polynomials. They can be expressed in the form \( R(x) = \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) \) is not zero.

• They often appear in calculus problems related to ratios or divided quantities.
• It's crucial to properly evaluate both the numerator and the denominator at a specific point before attempting to simplify the expression.
• If the denominator evaluates to zero at a point, it indicates the function may not be defined there.

In our exercise, the function \( \left(\frac{f}{g}\right)(x) \) is a rational function where we take \( f(x) \) as the numerator and \( g(x) \) as the denominator. To evaluate \( \left(\frac{f}{g}\right)(1) \), we evaluate \( f(1) \) and \( g(1) \) separately, then divide as \( \frac{0}{2} = 0 \). Understanding this concept is essential when dealing with mathematical expressions involving division or quotients.

The substitution method is widely used in algebra and calculus to simplify complex expressions by replacing a variable with its value or another expression. This approach allows you to focus on simpler calculations or analyses per component.

• Identify the variable to replace.
• Substitute it with the given value or another expression.
• Solve the resulting expression step-by-step.

In this problem, we applied the substitution method by replacing \( x \) with 1 in both \( f(x) \) and \( g(x) \). For \( f(x) = 2x^2 - 3x + 1 \), after substitution, we calculated each term: \( 2(1)^2 \), \( -3(1) \), and added 1. This gave us \( f(1) = 0 \). For \( g(x) = \frac{x^2 + 1}{x} \), substitution produced a straightforward calculation: \( g(1) = 2 \). Mastery of the substitution method is a key technique that simplifies solving and evaluating functions in calculus.