Math  /  Algebra

QuestionGiven the following functions: - f(x)=x22xf(x)=x^{2}-2 x - g(x)=x+3g(x)=\sqrt{x}+3
Determine fg(x)f \circ g(x).

Studdy Solution

STEP 1

1. We are given two functions, f(x)=x22x f(x) = x^2 - 2x and g(x)=x+3 g(x) = \sqrt{x} + 3 .
2. We need to find the composition of these functions, denoted as fg(x) f \circ g(x) , which means f(g(x)) f(g(x)) .

STEP 2

1. Understand the composition of functions.
2. Substitute g(x) g(x) into f(x) f(x) .
3. Simplify the expression.

STEP 3

The composition of functions fg(x) f \circ g(x) means we take the output of g(x) g(x) and use it as the input for f(x) f(x) . In other words, we need to evaluate f(g(x)) f(g(x)) .

STEP 4

Substitute g(x)=x+3 g(x) = \sqrt{x} + 3 into the function f(x)=x22x f(x) = x^2 - 2x . This means replacing every instance of x x in f(x) f(x) with g(x) g(x) .
f(g(x))=f(x+3)=(x+3)22(x+3) f(g(x)) = f(\sqrt{x} + 3) = (\sqrt{x} + 3)^2 - 2(\sqrt{x} + 3)

STEP 5

Expand the expression (x+3)2 (\sqrt{x} + 3)^2 :
(x+3)2=(x)2+2x3+32 (\sqrt{x} + 3)^2 = (\sqrt{x})^2 + 2 \cdot \sqrt{x} \cdot 3 + 3^2 =x+6x+9 = x + 6\sqrt{x} + 9

STEP 6

Substitute the expanded expression back into the function and simplify:
f(g(x))=x+6x+92(x+3) f(g(x)) = x + 6\sqrt{x} + 9 - 2(\sqrt{x} + 3)
Distribute the 2-2 in the second term:
=x+6x+92x6 = x + 6\sqrt{x} + 9 - 2\sqrt{x} - 6
Combine like terms:
=x+(6x2x)+(96) = x + (6\sqrt{x} - 2\sqrt{x}) + (9 - 6) =x+4x+3 = x + 4\sqrt{x} + 3
The composition fg(x) f \circ g(x) is x+4x+3 \boxed{x + 4\sqrt{x} + 3} .

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