Problem 21 - Precalculus : Building Concepts and Connections

Question

In Exercises \(17-40,\) let \(f(x)=-x^{2}+x, g(x)=\frac{2}{x+1},\) and \(h(x)=-2 x+1 .\) Evaluate each of the following. $$(f+h)(-2)$$

Step-by-Step Solution

Answer

(f+h)(-2) = -1

Step 1: Find the expression for f + h

We start off by adding the functions \(f\) and \(h\) together. The functions are \(f(x) = -x^{2}+x\) and \(h(x) = -2x+1\). Thus, the function \(f+h\) is given by: \((f+h)(x) = (-x^{2}+x) + (-2x+1) = -x^{2} - x + 1\).

Step 2: Compute the value at x=-2

Now we need to substitute the given value \(x = -2\) into our expression for \(f+h.\) So, we find \((f + h)(-2) = -(-2)^{2} - (-2) + 1 = -4 + 2 + 1 = -1\).

Key Concepts

Function Addition

When studying precalculus, understanding how to add two functions together is essential. Function addition is simply the process of combining two functions by adding their corresponding outputs for the same input value. For example, if you have two functions, say f(x) and g(x), the sum of these functions, denoted as (f + g)(x), is found by adding f(x) to g(x).

Here's a straightforward case: if f(x) = 2x and g(x) = 3x - 5, the function (f + g)(x) would be 2x + (3x - 5), which simplifies to 5x - 5. Notice that the operation is performed for each term that depends on the variable x.

Applying this to our exercise, the addition of f(x) and h(x), gave us a new function where each term of f(x) and h(x) that contained x was combined. By understanding and applying function addition, we can simplify the operations needed to evaluate the sum at any specific input.

Evaluating Functions

Evaluating functions is a fundamental skill in precalculus. To evaluate a function means to find the precise output when a specific input is plugged into the function's formula. Essentially, it involves replacing the variable x with the given number and then simplifying the expression using arithmetic operations.

To make this clearer, if you have a function f(x) = x^2 - 3x + 2 and you want to evaluate this function at x = 4, you'd replace every instance of x with 4: f(4) = (4)^2 - 3(4) + 2, which simplifies to 16 - 12 + 2 = 6.

In our textbook problem, evaluating the function (f + h)(-2) entailed substituting x with -2 into the combined function and finding the resulting value. This step can be tricky when dealing with negative numbers, so care should be taken to correctly apply the arithmetic operations to obtain the correct evaluation, (f + h)(-2) = -1.

Composition of Functions

The composition of functions is more complex than function addition. This concept involves applying one function to the result of another function. It's like a chain reaction where the output from the first function becomes the input for the second function.

To denote composition, we use the symbol \((g \circ f)(x)\), which means 'first apply \em f(x)\, then apply \em g\ to the result of \em f(x)\.' For example, if \em f(x) = x + 1\ and \em g(x) = x^2\, then the composition \((g \circ f)(x)\) is \em g(f(x))\, or specifically, \((x + 1)^2\). So, for any input \em x\, you first add 1 to \em x\, and then you square the result.

Although our original exercise did not require the use of composition, understanding this concept can vastly enhance a student’s toolkit when faced with complex functions. Mastery of function composition requires practice, as it often involves multiple steps and careful substitution to ensure success.