To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\).
We have: \[ f(x) = x^2 + 5x + 6 \] \[ g(x) = \frac{1}{x + 3} \]
Now we compute \((f \circ g)(x) = f(g(x))\): \[ f(g(x)) = f\left(\frac{1}{x + 3}\right) \]
Substituting \(\frac{1}{x + 3}\) into the function \(f(x)\): \[ f\left(\frac{1}{x + 3}\right) = \left(\frac{1}{x + 3}\right)^2 + 5\left(\frac{1}{x + 3}\right) + 6 \]
Now, we calculate each part:
- \(\left(\frac{1}{x + 3}\right)^2 = \frac{1}{(x + 3)^2}\)
- \(5\cdot\frac{1}{x + 3} = \frac{5}{x + 3}\)
- Adding the constant \(6\) can be expressed with a common denominator: \[ 6 = \frac{6(x + 3)^2}{(x + 3)^2} \]
Now we combine the terms: \[ f\left(\frac{1}{x + 3}\right) = \frac{1}{(x + 3)^2} + \frac{5}{x + 3} + \frac{6(x + 3)^2}{(x + 3)^2} \]
Finding a common denominator, which is \((x + 3)^2\): \[ = \frac{1 + 5(x + 3) + 6(x + 3)^2}{(x + 3)^2} \]
Now we can expand the numerator:
- \(1 + 5(x + 3) = 1 + 5x + 15 = 5x + 16\)
- \(6(x + 3)^2 = 6(x^2 + 6x + 9) = 6x^2 + 36x + 54\)
Combine the terms: \[ = 6x^2 + (5x + 36x) + (16 + 54) = 6x^2 + 41x + 70 \]
So we have: \[ f(g(x)) = \frac{6x^2 + 41x + 70}{(x + 3)^2} \]
Now combining everything: \[ f(g(x)) = \frac{6x^2 + 41x + 70}{x^2 + 6x + 9} \]
Therefore, the answer is: \[ \frac{6x^2 + 41x + 70}{x^2 + 6x + 9} \]
Thus, the correct response is: Start Fraction 6 x squared plus 41 x plus 70 over x squared plus 6 x plus 9 End Fraction.
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