Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 11-20: Find (a) \((f \circ g)(x)\) (b) \((g \circ f)(x)\) (c) \(f(g(-2))\) (d) \(g(f(3))\) $$ f(x)=2 x-5, \quad g(x)=3 x+7 $$

Step-by-Step Solution

Answer

(f ∘ g)(x) = 6x + 9, (g ∘ f)(x) = 6x - 8, f(g(-2)) = -3, g(f(3)) = 10.

Step 1: Understand Function Composition

Function composition means plugging one function into another. For the functions \(f(x) = 2x - 5\) and \(g(x) = 3x + 7\), to find \((f \circ g)(x)\), you substitute \(g(x)\) into \(f(x)\). This is written as \(f(g(x))\). Conversely, \((g \circ f)(x)\) is \(g(f(x))\).

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

Plug \(g(x) = 3x + 7\) into \(f(x) = 2x - 5\) to find \(f(g(x))\):\[f(g(x)) = f(3x + 7) = 2(3x + 7) - 5\]Simplify the expression:\[= 6x + 14 - 5 = 6x + 9\]

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

Plug \(f(x) = 2x - 5\) into \(g(x) = 3x + 7\) to find \(g(f(x))\):\[g(f(x)) = g(2x - 5) = 3(2x - 5) + 7\]Simplify the expression:\[= 6x - 15 + 7 = 6x - 8\]

Step 4: Calculate \(f(g(-2))\)

First, find \(g(-2)\):\[g(-2) = 3(-2) + 7 = -6 + 7 = 1\]Now, substitute 1 into \(f(x)\):\[f(1) = 2(1) - 5 = 2 - 5 = -3\]

Step 5: Calculate \(g(f(3))\)

First, find \(f(3)\):\[f(3) = 2(3) - 5 = 6 - 5 = 1\]Now, substitute 1 into \(g(x)\):\[g(1) = 3(1) + 7 = 3 + 7 = 10\]

Key Concepts

Composite Functions

Function composition is a fancy term, but don't let it intimidate you. It's simply about taking one function and inserting it into another. Imagine it like nesting functions. You have a function f(x) and another function g(x). When you compose them, you get new functions like \((f \circ g)(x)\) or \((g \circ f)(x)\). Here, \((f \circ g)(x)\) means you put g(x) inside f(x), and \((g \circ f)(x)\) means you put f(x) inside g(x).

In our exercise, first find \(f(g(x))\). Substitute g(x) into f(x): \(f(g(x)) = f(3x + 7)\). This means you replace every x in f(x) with g(x). Next, find \(g(f(x))\), which means put f(x) into g(x): \(g(f(x)) = g(2x - 5)\). Step by step, you follow these substitutions. With practice, this can actually become quite intuitive.

Algebraic Manipulation

Algebraic manipulation is at the heart of working with composite functions. This involves using basic algebra skills to simplify expressions or equations.

Let's break it down. When substituting g(x) into f(x), you start with \(f(g(x)) = 2(3x + 7) - 5\). Distribute the 2 into \(3x + 7\):

• First, multiply: \(2 \times 3x = 6x\)
• Then, \(2 \times 7 = 14\)
• Finally, subtract 5 from their sum: \(6x + 14 - 5 = 6x + 9\)

In the same fashion, for \(g(f(x)) = 3(2x - 5) + 7\), distribute the 3:

• \(3 \times 2x = 6x\)
• \(3 \times -5 = -15\)
• Add 7 to their sum: \(6x - 15 + 7 = 6x - 8\)

Through simple arithmetic, you end up with a neat result, neatly simplifying complex expressions.

Evaluating Functions

Evaluating functions is a method of determining the output of a function for a specific input. To do this, simply substitute the input values directly into the function.

Take the example of finding \(f(g(-2))\). Start by evaluating \(g(-2)\): you replace x in g(x) by -2 to get \(g(-2) = 3(-2) + 7 = -6 + 7 = 1\). Now, with \(g(-2) = 1\), evaluate f(1): substitute 1 into f(x) to find \(f(1) = 2(1) - 5 = 2 - 5 = -3\).

Similarly, evaluate \(g(f(3))\). First, find \(f(3)\) by putting 3 into f(x): \(f(3) = 2(3) - 5 = 6 - 5 = 1\). Following this, substitute into g(x): \(g(1) = 3(1) + 7 = 3 + 7 = 10\). Evaluating functions like this helps in breaking down what happens when specific values enter composite function combinations.