Problem 61 - 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. $$(g \circ f)(-3)$$

Step-by-Step Solution

Answer

The output of the function composition \( (g \circ f)(-3) \) is \( \sqrt{6} \).

Step 1: Evaluate f(-3)

First evaluate the function \( f(-3) \). \( f(x) = x^{2} + x \), thus, \( f(-3) = (-3)^{2} + (-3) = 9 - 3 = 6 \). So, \( f(-3) = 6 \).

Step 2: Evaluate g(f(-3))

Then, use the result from Step 1 to evaluate \( g(f(-3)) \). Since \( g(x) = \sqrt{x} \), this means that \( g(6) = \sqrt{6} \). Thus, \( (g \circ f)(-3) = g(6) = \sqrt{6} \).

Key Concepts

Composite Function

Understanding a composite function is like learning to apply a recipe step by step. When we talk about composite functions, we mean combining two functions in such a way that the output of one function becomes the input of the next. The notation \(g \circ f)(x)\) symbolizes this process and is read as 'g of f of x.' It's as if you're first passing your input, x, through the function f, and then taking that result and passing it through function g.

Imagine f as a machine that shapes clay and g as a machine that paints it. The clay (x) first goes through f, getting shaped, and then with its new form (f(x)), goes through g to get painted, giving us a beautifully finished product (g(f(x))). In our exercise, the shaping machine \(f\) squared the input and added that same input, while the painting machine \(g\) took the square root. Each machine must complete its job before the next starts.

Function Evaluation

Evaluating a function is like finding the exact amount of ingredients needed for a single serving of that recipe. You'd substitute the value of a single serving's worth of each ingredient into the recipe formula. Similarly, in mathematics, we substitute the given input value into the function's formula to find the output.

In the provided exercise, evaluating the function meant replacing the variable x with the actual input (-3) in the function \(f\). We calculated \(f(-3)\), which in our analogy is like measuring out the ingredients specifically for a serving size of -3. The resulting value was then used as the input for the next part of the recipe—the function g, which painted our shaped clay with the result of the shaping step.

Square Root Function

A square root function acts as a reverse process to squaring a number. If you envision a square with sides of length x, then \(x^2\) is the area, and \(\sqrt{x}\) is the length of a side. The square root function, denoted \(g(x)=\sqrt{x}\), extracts the original base number that was squared.

Considering a function machine analogy, think of the square root function as a machine that deconstructs a building block into its original form. In our exercise, once we had the value from the quadratic function \(f\), we fed it into our square root function \(g\), which essentially asked, 'What number, when squared, gives me this value?' The machine then disassembled the building block until it was back to its base form, giving us \(\sqrt{6}\) for the input of 6.

Quadratic Function

The quadratic function is an essential concept in algebra, and it can be visualized as a curved parabola when graphed. In the general form, a quadratic function is expressed as \(f(x) = ax^2 + bx + c\), where 'a' is not zero. This type of function is characterized by the highest exponent of x being 2, hence 'quad' for 'square'.

The quadratic function can model various phenomena such as projectile motion or the shape of a satellite dish. In our exercise, the quadratic function \(f\) took the input -3, squared it which resulted in 9, and added the original input to this result, ultimately giving us 6. It's important to understand this function because it represents a fundamental relationship between an input and its square, setting the stage for the composite processes we explore in such exercises.