Q1: Find f + g, f − g, fg, and f/g and their domains.
f(x) = √16 − x2, g(x) = √x + 3
a) Find (f + g)(x).
b) Find the domain of (f + g)(x).(Enter your answer using interval notation.)
c) Find (f − g)(x).
d) Find the domain of (f − g)(x).(Enter your answer using interval notation.)
e) Find (fg)(x).
f) Find the domain of (fg)(x).(Enter your answer using interval notation.)
g) Find (f/g)(x).
h) Find the domain of (f/g)(x).(Enter your answer using interval notation.)
Trying to deal with your lack of parentheses, I get
(f/g)(x) = f(x)/g(x) = √(16-x^2)/√(x+3)
The domain excludes x <= -3 because of the denominator
and also |x| > 4 due to the numerator. So that leaves (-3,4]
See what you can do with the rest; they are similar.
a) (f + g)(x) = √16 - x^2 + √x + 3
b) The domain of (f + g)(x) is the intersection of the domains of f(x) and g(x).
For f(x), the square root function is defined for non-negative values of its argument. So, 16 - x^2 ≥ 0. Solving for x, we get -4 ≤ x ≤ 4.
For g(x), the square root function is always defined for non-negative values, so x + 3 ≥ 0. Solving for x, we get x ≥ -3.
Therefore, the domain of (f + g)(x) is the intersection of -4 ≤ x ≤ 4 and x ≥ -3, which is -3 ≤ x ≤ 4.
c) (f - g)(x) = √16 - x^2 - √x + 3
d) Similar to part b, the domain of (f - g)(x) is the intersection of the domains of f(x) and g(x).
For f(x), -4 ≤ x ≤ 4.
For g(x), x + 3 ≥ 0. So, x ≥ -3.
Therefore, the domain of (f - g)(x) is the intersection of -4 ≤ x ≤ 4 and x ≥ -3, which is -3 ≤ x ≤ 4.
e) (fg)(x) = (√16 - x^2)(√x + 3)
f) The domain of (fg)(x) is the intersection of the domains of f(x) and g(x).
For f(x), -4 ≤ x ≤ 4.
For g(x), √x + 3 is defined for x ≥ -3.
Therefore, the domain of (fg)(x) is the intersection of -4 ≤ x ≤ 4 and x ≥ -3, which is -3 ≤ x ≤ 4.
g) (f/g)(x) = (√16 - x^2)/(√x + 3)
h) The domain of (f/g)(x) is the intersection of the domains of f(x) and g(x).
For f(x), -4 ≤ x ≤ 4.
For g(x), √x + 3 is defined for x ≥ -3.
However, since we have a division by √x + 3, we need to make sure that √x + 3 ≠ 0, which means x ≠ -3.
Therefore, the domain of (f/g)(x) is -4 ≤ x ≤ 4, x ≠ -3.
To find f + g, f - g, fg, and f/g, as well as their domains, we need to perform the given operations on the functions f(x) = √16 - x^2 and g(x) = √x + 3. Let's go through each step.
a) To find (f + g)(x), we need to add the two functions:
(f + g)(x) = f(x) + g(x) = √16 - x^2 + √x + 3
b) To determine the domain of (f + g)(x), we need to consider the domains of f(x) and g(x) and find the intersection.
The domain is restricted by two conditions:
- The expression under the square root (√) should be non-negative (i.e., greater than or equal to zero).
- The denominator of a fraction cannot be zero.
Let's examine each function's domain:
- For f(x) = √16 - x^2, the expression under the square root should be greater than or equal to zero.
16 - x^2 ≥ 0
Solving this inequality, we have x ≤ 4 and x ≥ -4. So the domain of f(x) is [-4, 4].
- For g(x) = √x + 3, the expression under the square root should be greater than or equal to zero.
x + 3 ≥ 0
Solving this inequality, we have x ≥ -3. So the domain of g(x) is [-3, ∞).
The domain of (f + g)(x) is the intersection of the domains of f(x) and g(x), which is [-3, 4].
c) To find (f - g)(x), we need to subtract the second function from the first:
(f - g)(x) = f(x) - g(x) = √16 - x^2 - √x + 3
d) To determine the domain of (f - g)(x), we apply the same rules as in part b.
The domain is given by the intersection of the domains of f(x) and g(x), which is [-3, 4].
e) To find (fg)(x), we need to multiply the two functions:
(fg)(x) = f(x) * g(x) = (√16 - x^2) * (√x + 3)
f) To determine the domain of (fg)(x), we apply the same rules as in part b.
The domain is given by the intersection of the domains of f(x) and g(x), which is [-3, 4].
g) To find (f/g)(x), we need to divide the first function by the second:
(f/g)(x) = f(x) / g(x) = (√16 - x^2) / (√x + 3)
h) To determine the domain of (f/g)(x), we need to consider the domain restrictions:
- The expression under the square root (√) should be non-negative (i.e., greater than or equal to zero).
- The denominator of a fraction cannot be zero.
Let's examine each function's domain again:
- For f(x) = √16 - x^2, the domain is [-4, 4].
- For g(x) = √x + 3, the expression under the square root should be greater than or equal to zero.
x + 3 ≥ 0
Solving this inequality, we have x ≥ -3. So the domain of g(x) is [-3, ∞).
The domain of (f/g)(x) is the intersection of the domains of f(x) and g(x), which is [-4, 4].