QUESTION IMAGE
Question
for the functions f(x)=5 - x^2 and g(x)=x^2 + 6x - 27, find f + g, f - g, fg, and f/g. determine the domain for each function. b. the domain of f - g is (-∞,∞). (type your answer in interval notation.) c. the domain of f - g is ∅. (f·g)(x)= (simplify your answer.) what is the domain of fg? select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the domain of fg is { }. (use a comma to separate answers as needed.) b. the domain of fg is. (type your answer in interval notation.) c. the domain of fg is ∅.
Step1: Find (f + g)(x)
(f + g)(x)=f(x)+g(x)=(5 - x^{2})+(x^{2}+6x - 27)=6x - 22. The domain of a linear - function is all real numbers, so the domain of f + g is $(-\infty,\infty)$.
Step2: Find (f - g)(x)
(f - g)(x)=f(x)-g(x)=(5 - x^{2})-(x^{2}+6x - 27)=5 - x^{2}-x^{2}-6x + 27=-2x^{2}-6x + 32. The domain of a quadratic - function is all real numbers, so the domain of f - g is $(-\infty,\infty)$.
Step3: Find (f\cdot g)(x)
(f\cdot g)(x)=f(x)\cdot g(x)=(5 - x^{2})(x^{2}+6x - 27)
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The domain of a polynomial function is all real numbers, so the domain of f\cdot g is $(-\infty,\infty)$.
Step4: Find $\frac{f}{g}(x)$
$\frac{f}{g}(x)=\frac{5 - x^{2}}{x^{2}+6x - 27}$. The denominator cannot be zero. Set $x^{2}+6x - 27 = 0$. Factor the quadratic equation: $x^{2}+6x - 27=(x + 9)(x - 3)=0$. Solving $(x + 9)(x - 3)=0$ gives $x=-9$ or $x = 3$. So the domain of $\frac{f}{g}$ is $(-\infty,-9)\cup(-9,3)\cup(3,\infty)$.
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- $(f + g)(x)=6x - 22$, domain: $(-\infty,\infty)$
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