Question 2.7.2: If f(x)=√4-x² and g(x)=3x+1, find (f+g)(x),(f-g)(x),(fg)(x),......
If f(x)=\sqrt {4-x²} and g(x)=3x+1, find (f+g)(x),(f-g)(x),(fg)(x), and (f/g)(x), and state the domains of the respective functions.
f(x) = √(4-x²)
g(x) = 3x+1
Step 1:
To determine the domain of the functions f and g, we need to consider any restrictions on the values of x that would make the functions undefined.
Step 2:
For f(x), the square root function is defined for all non-negative real numbers. Therefore, the expression inside the square root, 4-x^2, must be greater than or equal to 0. Solving this inequality, we find that -2 ≤ x ≤ 2. So, the domain of f is the closed interval [-2, 2].
Step 3:
For g(x), there are no restrictions on the domain since it is defined for all real numbers. Therefore, the domain of g is the set of all real numbers, denoted as ℝ.
Step 4:
To find the domain of f+g, f-g, and fg, we need to consider the intersection of the domains of f and g. Since the domain of f is [-2, 2] and the domain of g is ℝ, their intersection is [-2, 2].
Step 5:
For the domain of f/g, we need to exclude any values of x that would make the denominator, 3x+1, equal to 0. Solving this equation, we find that x = -1/3. Therefore, the domain of f/g is the interval [-2, 2], excluding x = -1/3.
Final Answer
The domain of f is the closed interval [-2, 2], and the domain of g is \mathbb{R}. The intersection of these domains is [-2, 2], which is the domain of f+g, f-g, and fg. For the domain of f/g, we exclude each number x in [-2, 2] such that g(x)=3 x+1=0( namely, \left.x=-\frac{1}{3}\right). Thus, we have the following:
\begin{aligned}(f+g)(x) & =\sqrt{4-x^2}+(3 x+1), & & -2 \leq x \leq 2 \\(f-g)(x) & =\sqrt{4-x^2}-(3 x+1), & & -2 \leq x \leq 2 \\(f g)(x) & =\sqrt{4-x^2}(3 x+1), & & -2 \leq x \leq 2 \\\left(\frac{f}{g}\right)(x) & =\frac{\sqrt{4-x^2}}{3 x+1}, & & -2 \leq x \leq 2 \text { and } x \neq-\frac{1}{3}\end{aligned}
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