Problem 50 - Precalculus : Building Concepts and Connections

Question

In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(f \circ h)(1)$$

Step-by-Step Solution

Answer

The value of the composite function (f o h)(1) is 6.

Step 1: Evaluate Inner Function

First, evaluate the inner function h at x = 1. We have h(x) = -3x, therefore h(1) = -3*1 which equals -3.

Step 2: Evaluate Composite Function

Next, substitute the result from Step 1 into the function f. The function f(x) is given by x^2 + x. Substituting x = -3 into f(x) gives f(-3) = (-3)^2 + -3 = 9 - 3 = 6.

Key Concepts

Function Composition

Function composition is a fundamental concept in precalculus involving combining two or more functions to create a new function. Think of it as a sequence of operations: you perform one function, take the result, and then feed it into the next function. In the given exercise, we're tasked with finding \( (f \circ h)(1) \) — this reads as 'f composed with h of 1'. Here's how it works:

First, you take the function inside the parentheses, \( h(x) = -3x \), and evaluate it at \( x = 1 \), obtaining \( h(1) = -3 \). This value of \( -3 \) is then used as the input for the function \( f \), which follows the composition symbol \( \circ \). By substituting \( -3 \) into \( f(x) = x^2 + x \) you get \( f(-3) \) which yields the final result. This chained operation showcases the essence of function composition, a powerful tool in advanced mathematics.

Evaluating Functions

Evaluating functions involves finding the output of a function for a given input value. It's like solving a puzzle: the function provides the rules, and you plug in the missing piece - the input - to reveal the complete picture - the output. When evaluating \( h(x) \), we input \( x = 1 \) to obtain \( h(1) = -3x \), leading us to \( -3 \) when we substitute \( 1 \) for \( x \). The same step-by-step approach applies to the function \( f(x) \) when we use the output of \( h(1) \) as the new input. The key here is to carefully substitute and simplify at each step, ensuring that no errors are made in the arithmetic operations.

Precalculus Problem Solving

Precalculus problem solving demands a logical and methodical approach to dealing with mathematical challenges. Tackling problems like function composition involves a clear understanding of each individual function before attempting to combine them. In the exercise we're working through, breaking down the problem into manageable steps is crucial. We first handled \( h(1) \), followed by \( f(h(1)) \), ensuring a coherent flow in calculations. Attributes like critical thinking and attention to detail are invaluable here, as they prevent common mistakes that can occur when handling complex expressions or negative numbers, as seen with the \( f(-3) \) step.

Operations on Functions

Operations on functions are not limited to composition. They include addition, subtraction, multiplication, and division of functions, similar to operations on numbers. Each operation has its own set of rules and applications. In composition, as seen in this exercise, we 'feed' the output of one function into another. But in other operations, we might be required to perform function addition, which involves adding the values of two functions at the same point. For instance, \( f(x) + g(x) \) would entail adding together the solutions to each function for the same value of \( x \). Understanding the various operations available can greatly enhance problem-solving techniques in precalculus.