use f(x)=square root of 5-x and g(x)=x^2 + 2x
a-find f(g(x)), state the domain
b-find g(f(x)),state the domain
c-find g(f(-1))
d-find f(x)/g(x), state the domain, then evaluate for x=3
f(x) = √(5-x) , g(x) = x^2 + 2x
f(g(x)
= √(5 - (x^2+2x) )
= √(5 - x^2 - 2x)
for this to be real, 5 - x^2 - 2x ≥ 0
x^2 + 2x - 5 ≤ 0
solve x^2 + 2x - 5 = 0 ,
the domain will be all values of x between the two solution values
g(f(x)
= (√(5-x))^2 + 2√(5-x)
= 5 - x + 2√(5-x)
domain is :
5-x ≥ 0
-x ≥ -5
x ≤ 5
you try c and d
To find f(g(x)), we need to substitute g(x) into f(x):
a) f(g(x)) = sqrt(5 - (x^2 + 2x))
To simplify, first simplify the expression inside the square root:
= sqrt(5 - x^2 - 2x)
The domain of f(x) is restricted by the domain of the square root function, which requires that the expression inside the square root is non-negative. So, to find the domain of f(g(x)), we need to solve the inequality:
5 - x^2 - 2x ≥ 0
To solve this inequality, we can factor the quadratic expression and solve for x:
(1-x)(5+x) ≥ 0
Now, we have two cases to consider:
1) Case 1: 1-x ≥ 0 and 5+x ≥ 0
Solve both inequalities separately:
-x ≥ -1 (multiplying both sides by -1, which reverses the inequality)
x ≤ 1 (dividing both sides by -1, which reverses the inequality)
x ≥ -5 (subtracting 5 from both sides)
The common interval is -5 ≤ x ≤ 1.
2) Case 2: 1-x ≤ 0 and 5+x ≤ 0
Solve both inequalities separately:
-x ≤ 1
x ≥ -1
x ≤ -5
The common interval is -5 ≤ x ≤ -1.
Therefore, the domain of f(g(x)) is -5 ≤ x ≤ 1.
b) To find g(f(x)), substitute f(x) into g(x):
g(f(x)) = (sqrt(5 - x))^2 + 2(sqrt(5 - x))
= 5 - x + 2(sqrt(5 - x))
The domain of sqrt(5 - x) is restricted by the requirement that the expression inside the square root is non-negative. So, the domain of g(f(x)) is also determined by the domain of the square root function, which is 5 - x ≥ 0. Solving this inequality:
5 - x ≥ 0
x ≤ 5
Therefore, the domain of g(f(x)) is x ≤ 5.
c) To find g(f(-1)), substitute -1 into f(x) and then substitute the result into g(x):
f(-1) = sqrt(5 - (-1))
= sqrt(6)
g(f(-1)) = (sqrt(6))^2 + 2(sqrt(6))
= 6 + 2(sqrt(6))
So, g(f(-1)) = 6 + 2(sqrt(6)).
d) To find f(x)/g(x), substitute the expressions of f(x) and g(x):
f(x)/g(x) = sqrt(5 - x) / (x^2 + 2x)
The domain of f(x)/g(x) is restricted by the requirement that the denominator is not equal to zero. We need to solve the equation:
x^2 + 2x ≠ 0
The expression is a quadratic function, and to solve it, we factor it:
x(x + 2) ≠ 0
The solutions to this equation are x ≠ 0 and x ≠ -2.
Therefore, the domain of f(x)/g(x) is all real numbers except x = 0 and x = -2.
To evaluate f(x)/g(x) for x = 3, substitute 3 into the expression:
f(3)/g(3) = sqrt(5 - 3) / (3^2 + 2(3))
= sqrt(2) / (9 + 6)
= sqrt(2) / 15