Math  /  Algebra

Questionthe functions f(x)=2x+3f(x)=\frac{2}{x+3} and g(x)=11x+2g(x)=\frac{11}{x+2}, find the compositio tation. (fg)(x)=(f \circ g)(x)= \square
Domain of fgf \circ g : \square

Studdy Solution

STEP 1

1. The composition of functions (fg)(x) (f \circ g)(x) is defined as f(g(x)) f(g(x)) .
2. The domain of a composition fg f \circ g is determined by the domain of g(x) g(x) and the domain of f(x) f(x) applied to the range of g(x) g(x) .

STEP 2

1. Find the expression for (fg)(x) (f \circ g)(x) .
2. Determine the domain of g(x) g(x) .
3. Determine the domain of f(x) f(x) applied to g(x) g(x) .
4. Find the domain of the composition fg f \circ g .

STEP 3

To find (fg)(x) (f \circ g)(x) , substitute g(x) g(x) into f(x) f(x) :
f(g(x))=f(11x+2)=211x+2+3 f(g(x)) = f\left(\frac{11}{x+2}\right) = \frac{2}{\frac{11}{x+2} + 3}
Simplify the expression:
=211x+2+3(x+2)x+2 = \frac{2}{\frac{11}{x+2} + \frac{3(x+2)}{x+2}}
=211+3(x+2)x+2 = \frac{2}{\frac{11 + 3(x+2)}{x+2}}
=2(x+2)11+3(x+2) = \frac{2(x+2)}{11 + 3(x+2)}
=2(x+2)11+3x+6 = \frac{2(x+2)}{11 + 3x + 6}
=2(x+2)3x+17 = \frac{2(x+2)}{3x + 17}

STEP 4

Determine the domain of g(x)=11x+2 g(x) = \frac{11}{x+2} . The function g(x) g(x) is undefined when its denominator is zero:
x+20 x + 2 \neq 0
x2 x \neq -2

STEP 5

Determine where f(x)=2x+3 f(x) = \frac{2}{x+3} is defined when x=g(x) x = g(x) . The function f(x) f(x) is undefined when its denominator is zero:
11x+2+30 \frac{11}{x+2} + 3 \neq 0
11+3(x+2)x+20 \frac{11 + 3(x+2)}{x+2} \neq 0
11+3x+60 11 + 3x + 6 \neq 0
3x+170 3x + 17 \neq 0
x173 x \neq -\frac{17}{3}

STEP 6

The domain of fg f \circ g is the set of x x values that satisfy both domain restrictions:
1. x2 x \neq -2
2. x173 x \neq -\frac{17}{3}

Thus, the domain of fg f \circ g is all real numbers except x=2 x = -2 and x=173 x = -\frac{17}{3} .
The composition (fg)(x)=2(x+2)3x+17 (f \circ g)(x) = \frac{2(x+2)}{3x + 17} .
The domain of fg f \circ g is all real numbers x x such that x2 x \neq -2 and x173 x \neq -\frac{17}{3} .

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