Problem 18 - Contemporary Precalculus

Question

Find \((g \circ f)(3),(f \circ g)(1),\) and \((f \circ f)(0)\) $$f(x)=|x+2|, \quad g(x)=-x^{2}$$

Step-by-Step Solution

Answer

Question: Find the values of (g ∘ f)(3), (f ∘ g)(1), and (f ∘ f)(0) for the functions f(x) = |x+2| and g(x) = -x^2. Answer: The values of the compositions are (g ∘ f)(3) = -25, (f ∘ g)(1) = 1, and (f ∘ f)(0) = 4.

Step 1: Calculate \((g \circ f)(3)\)

\ To find \((g \circ f)(3)\), we have to first find the value of \(f(3)\) and then use this value as input to the function \(g\). So, let's first determine \(f(3)\). $$f(3)=|3+2|=|5|=5$$ Now, substitute the value of \(f(3)\) into the function \(g\). $$(g \circ f)(3)=g(f(3))=g(5)=-5^{2}=-25$$

Step 2: Calculate \((f \circ g)(1)\)

\ To find \((f \circ g)(1)\), we need to first find the value of \(g(1)\) and then use this value as input to the function \(f\). Let's find \(g(1)\). $$g(1)=-1^{2}=-1$$ Now, substitute the value of \(g(1)\) into the function \(f\). $$(f \circ g)(1)=f(g(1))=f(-1)=|-1+2|=|1|=1$$

Step 3: Calculate \((f \circ f)(0)\)

\ To find \((f \circ f)(0)\), we need to first find the value of \(f(0)\) and then use this value as input to the function \(f\) again. Let's find \(f(0)\). $$f(0)=|0+2|=|2|=2$$ Now, substitute the value of \(f(0)\) into the function \(f\). $$(f \circ f)(0)=f(f(0))=f(2)=|2+2|=|4|=4$$ The requested values are: $$(g \circ f)(3) = -25$$ $$(f \circ g)(1) = 1$$ $$(f \circ f)(0) = 4$$

Key Concepts

Function Operations

Understanding function operations is crucial when learning mathematics. When we talk about function operations, we're referring to ways in which we can combine functions to create new ones. One of the principal operations is composition, where we apply one function to the result of another.

In the given exercise, we deal with function composition denoted as \( (g \circ f)(x) \). This notation means that we first apply the function \( f \) to \( x \), and then apply the function \( g \) to the result of that. It's important to follow the order of operations carefully – applying \( f \) first and then \( g \) can yield a different result than doing it the other way around, as illustrated by the results of \( (g \circ f)(3) \) and \( (f \circ g)(1) \) in the exercise provided.

Absolute Value Function

The absolute value function is represented by \( |x| \), which essentially provides the distance of a number \( x \) from zero on the number line, disregarding the direction. Thus, it always returns a non-negative result. When dealing with absolute value functions like \( f(x) = | x + 2 | \) from our exercise, remember that the outcome of the function is the same whether \( x \) is positive or negative once it's inside the absolute value bars.

For example, to find \( f(3) \) we evaluated \( |3+2| \) and got 5, irrespective of the sign of 3. Similarly, for \( f(-1) \) the calculation of \( |-1+2| \) gave us 1. Understanding that the absolute value function 'strips' the number of its sign can help in simplifying the steps when solving absolute value equations or compositions involving them.

Quadratic Function

A quadratic function takes the form \( g(x) = ax^2 + bx + c \) where \( a \) is not zero. It is characterized by a parabola when graphed on a coordinate plane, with its direction depending on the sign of \( a \).

In our problem, the function \( g(x) = -x^2 \) is a simplified quadratic function. The negative sign in front of \( x^2 \) means that the parabola opens downwards. Quadratic functions are often involved in calculating areas, trajectories, and can represent a variety of physical phenomena.

For instance, when we calculated \( (g \circ f)(3) \), we plugged the output from the absolute value function, which was 5, into our quadratic function \( g \) resulting in \( -5^{2} = -25 \). This sign before the \( x^2 \) term is crucial as it changes the entire outcome of the function, indicating that understanding the properties of quadratic functions is essential for proper evaluation.