Find an equivalent algebraic expression for each composition.
The Correct Answer and Explanation is:
Problem:
Let
- f(x)=2x+3f(x) = 2x + 3f(x)=2x+3
- g(x)=x2−1g(x) = x^2 – 1g(x)=x2−1
Find the algebraic expression for (f∘g)(x)(f \circ g)(x)(f∘g)(x) and (g∘f)(x)(g \circ f)(x)(g∘f)(x).
Answer:
- (f∘g)(x)=f(g(x))=2(x2−1)+3=2×2−2+3=2×2+1(f \circ g)(x) = f(g(x)) = 2(x^2 – 1) + 3 = 2x^2 – 2 + 3 = 2x^2 + 1(f∘g)(x)=f(g(x))=2(x2−1)+3=2×2−2+3=2×2+1
- (g∘f)(x)=g(f(x))=(2x+3)2−1=4×2+12x+9−1=4×2+12x+8(g \circ f)(x) = g(f(x)) = (2x + 3)^2 – 1 = 4x^2 + 12x + 9 – 1 = 4x^2 + 12x + 8(g∘f)(x)=g(f(x))=(2x+3)2−1=4×2+12x+9−1=4×2+12x+8
Explanation
Function composition involves applying one function to the result of another. It is written as (f∘g)(x)(f \circ g)(x)(f∘g)(x), which means we apply g(x)g(x)g(x) first and then apply fff to that result. To understand function composition better, we can think of it like passing input through two machines in sequence. The output of the first machine becomes the input for the second.
Let’s start with (f∘g)(x)(f \circ g)(x)(f∘g)(x). This notation means “f of g of x.” We begin by calculating g(x)g(x)g(x), which is given as x2−1x^2 – 1×2−1. Then we plug that into the function f(x)=2x+3f(x) = 2x + 3f(x)=2x+3. Substituting g(x)g(x)g(x) into fff gives us f(g(x))=2(x2−1)+3f(g(x)) = 2(x^2 – 1) + 3f(g(x))=2(x2−1)+3. Distributing the 2 gives 2×2−2+32x^2 – 2 + 32×2−2+3, which simplifies to 2×2+12x^2 + 12×2+1.
Next, we find (g∘f)(x)(g \circ f)(x)(g∘f)(x). This is the reverse: “g of f of x.” First, we compute f(x)=2x+3f(x) = 2x + 3f(x)=2x+3 and plug that result into g(x)=x2−1g(x) = x^2 – 1g(x)=x2−1. So, g(f(x))=(2x+3)2−1g(f(x)) = (2x + 3)^2 – 1g(f(x))=(2x+3)2−1. Expanding (2x+3)2(2x + 3)^2(2x+3)2 results in 4×2+12x+94x^2 + 12x + 94×2+12x+9. Then subtracting 1 gives the final expression 4×2+12x+84x^2 + 12x + 84×2+12x+8.
These two expressions show that function composition is not generally commutative. That means f∘gf \circ gf∘g is not usually the same as g∘fg \circ fg∘f. It is important to carefully follow the order when composing functions to avoid mistakes.
