for the given functions f and g, find the following and state the domain of each.
f(x)=square root x; g(x)=5x-1
a. (f*g)(x)= ?
what is the domain of f*g?
b. (f/g)(x)=?
what is the domain of f/g?
please show work
domains:
f: x>=0
g: all real
f*g = f(x)*g(x) = (5x-1)√x
domain: x>=0
f/g = f(x)/g(x) = √x / (5x-1)
domain x>0 but x ≠ 1/5
That is, (0,1/5)U(1/5,oo)
Describe the domain of the function f (X) =x+4/x square_1 solve the problem
a. To find (f*g)(x), we need to perform the operation of composition, which means evaluating one function within another.
(f*g)(x) = f(g(x))
So, substituting g(x) into f(x):
(f*g)(x) = f(5x-1)
Now, we substitute the expression 5x-1 into the function f(x):
(f*g)(x) = square root (5x-1)
To find the domain of (f*g)(x), we need to determine the values of x for which the function is defined. Since we are taking the square root of a quantity (5x-1), the function is only defined for values of x that make 5x-1 greater than or equal to 0. In other words:
5x-1 >= 0
Solving this inequality, we get:
5x >= 1
x >= 1/5
Therefore, the domain of (f*g)(x) is x>=1/5.
b. To find (f/g)(x), we need to perform the operation of division between the two functions:
(f/g)(x) = f(x) / g(x)
Substituting the given functions into the division:
(f/g)(x) = square root x / (5x-1)
To find the domain of (f/g)(x), we need to determine the values of x for which the function is defined. Since division by zero is undefined, the function is defined for any value of x that does not make the denominator, (5x-1), equal to zero. In other words:
5x-1 != 0
Solving this inequality, we get:
5x != 1
x != 1/5
Therefore, the domain of (f/g)(x) is x != 1/5.
To find (f*g)(x), we need to multiply the two functions, f(x) and g(x).
a. (f*g)(x) = f(x) * g(x) = (sqrt(x)) * (5x-1)
To simplify this, we multiply the terms inside the brackets:
= sqrt(x) * 5x - sqrt(x) * 1
= 5x^(3/2) - sqrt(x)
The domain of f*g will be the set of all x-values for which the function is defined. In this case, both f(x) = sqrt(x) and g(x) = 5x-1 are defined for all real numbers. Therefore, the domain of f*g is the set of all real numbers: (-∞, ∞).
b. (f/g)(x) = f(x) / g(x) = (sqrt(x)) / (5x-1)
The division of two functions results in their quotient. The domain of f/g will be the set of all x-values that make the denominator non-zero. In this case, the denominator 5x-1 cannot be zero, so we solve the equation:
5x - 1 ≠ 0
5x ≠ 1
x ≠ 1/5
Therefore, the domain of f/g is all real numbers except x = 1/5. In interval notation, the domain is (-∞, 1/5) U (1/5, ∞).