f(x)= ¡Ìx+4, g(x)= 2/¡Ì7-2x
1. The domain of f is_______(in interval notation)
2. The domain of g is ______(in interval notation)
3. (f+g)(0)=
4. (f+g)(-5)=
To find the domain of a function, we need to identify the set of all possible inputs for which the function is defined.
1. The domain of f(x) can be calculated by considering the term inside the square root function, which is x+4. For a square root function to be defined, the expression inside the square root must be greater than or equal to zero. So we have:
x + 4 ≥ 0
Solving for x, we find:
x ≥ -4
Therefore, the domain of f(x) is [ -4, ∞ ) in interval notation.
2. The domain of g(x) can be determined by considering the term inside the square root function, which is 7 - 2x. Again, for a square root function to be defined, the expression inside the square root must be greater than or equal to zero. So we have:
7 - 2x ≥ 0
Solving for x, we find:
-2x ≥ -7
x ≤ 7/2
Therefore, the domain of g(x) is ( -∞, 7/2 ] in interval notation.
3. To find (f+g)(0), we need to add the functions f(x) and g(x) and then evaluate the resulting expression at x = 0.
(f+g)(x) = f(x) + g(x)
Substituting f(x) = √(x+4) and g(x) = 2/√(7-2x), we have:
(f+g)(x) = √(x+4) + 2/√(7-2x)
Now, plugging in x = 0, we get:
(f+g)(0) = √(0+4) + 2/√(7-2(0))
Simplifying further:
(f+g)(0) = √4 + 2/√7
Since √4 = 2, and we cannot simplify √7 any further, the final answer is:
(f+g)(0) = 2 + 2/√7
4. To find (f+g)(-5), we use the same process as in the previous question:
(f+g)(x) = f(x) + g(x)
Substituting f(x) = √(x+4) and g(x) = 2/√(7-2x), we have:
(f+g)(x) = √(x+4) + 2/√(7-2x)
Now, plugging in x = -5, we get:
(f+g)(-5) = √(-5+4) + 2/√(7-2(-5))
Simplifying further:
(f+g)(-5) = √(-1) + 2/√(7+10)
Notice that √(-1) is undefined since square root of a negative number is not defined in the set of real numbers. Therefore, (f+g)(-5) is undefined.