To find the value of (f∘g)(x), we need to substitute g(x) into f(x):
(f∘g)(x) = f(g(x)) = f (√(x-2))
Given that f(x) = 1 / (x^2 + 3), we substitute (√(x-2)) for x:
(f∘g)(x) = f(√(x-2)) = 1 / ((√(x-2))^2 + 3) = 1 / (x-2 + 3) = 1 / (x+1)
So, (f∘g)(x) = 1 / (x+1)
Find left-parenthesis lower f ringcircle lower g right-parenthesis left-parenthesis x right-parenthesis where lower f left-parenthesis x right-parenthesis equals Start Fraction 1 over x squared plus 3 End Fraction and lower g left-parenthesis x right-parenthesis equals Start Root x minus 2 End Root.
9 answers
Find left-parenthesis lower g circle lower f right-parenthesis left-parenthesis x right-parenthesis where lower f left-parenthesis x right-parenthesis equals x squared minus 2 and lower g left-parenthesis x right-parenthesis equals 5 x minus 8.
To find the value of (g∘f)(x), we need to substitute f(x) into g(x):
(g∘f)(x) = g(f(x)) = g(x^2 - 2)
Given that g(x) = 5x - 8, we substitute (x^2 - 2) for x:
(g∘f)(x) = g(x^2 - 2) = 5(x^2 - 2) - 8 = 5x^2 - 10 - 8 = 5x^2 - 18
So, (g∘f)(x) = 5x^2 - 18
(g∘f)(x) = g(f(x)) = g(x^2 - 2)
Given that g(x) = 5x - 8, we substitute (x^2 - 2) for x:
(g∘f)(x) = g(x^2 - 2) = 5(x^2 - 2) - 8 = 5x^2 - 10 - 8 = 5x^2 - 18
So, (g∘f)(x) = 5x^2 - 18
Determine the domain of the function left-parenthesis lower f circle lower g right-parenthesis left-parenthesis x right-parenthesis where lower f left-parenthesis x right-parenthesis equals Start Fraction x squared over x squared minus 1 End Fraction and lower g left-parenthesis x right-parenthesis equals Start Root x plus 4 End Root.
To determine the domain of the function (f∘g)(x), we need to consider the domains of both f(x) and g(x) and find the intersection.
Given that f(x) = x^2 / (x^2 - 1) and g(x) = √(x + 4), we need to find the domain of both functions.
For f(x), the denominator cannot be zero, so x^2 - 1 ≠ 0. Solving the equation, we get x ≠ ±1. Therefore, the domain of f(x) is all real numbers except x = ±1.
For g(x), the argument of the square root must be greater than or equal to zero, so x + 4 ≥ 0. Solving the inequality, we get x ≥ -4. Therefore, the domain of g(x) is x ≥ -4.
To find the domain of (f∘g)(x), we need to find the intersection of the domains of f(x) and g(x). Since the domains of f(x) and g(x) overlap for x ≥ -4, the domain of (f∘g)(x) is x ≥ -4, excluding x = ±1.
In interval notation, the domain of (f∘g)(x) is (-∞, -4) U (-4, 1) U (1, ∞).
Given that f(x) = x^2 / (x^2 - 1) and g(x) = √(x + 4), we need to find the domain of both functions.
For f(x), the denominator cannot be zero, so x^2 - 1 ≠ 0. Solving the equation, we get x ≠ ±1. Therefore, the domain of f(x) is all real numbers except x = ±1.
For g(x), the argument of the square root must be greater than or equal to zero, so x + 4 ≥ 0. Solving the inequality, we get x ≥ -4. Therefore, the domain of g(x) is x ≥ -4.
To find the domain of (f∘g)(x), we need to find the intersection of the domains of f(x) and g(x). Since the domains of f(x) and g(x) overlap for x ≥ -4, the domain of (f∘g)(x) is x ≥ -4, excluding x = ±1.
In interval notation, the domain of (f∘g)(x) is (-∞, -4) U (-4, 1) U (1, ∞).
choices
left-parenthesis negative infinity comma negative 1 right-parenthesis union left-parenthesis negative 1 comma 1 right-parenthesis union left-parenthesis 1 comma infinity right-parenthesisImage with alt text: left-parenthesis negative infinity comma negative 1 right-parenthesis union left-parenthesis negative 1 comma 1 right-parenthesis union left-parenthesis 1 comma infinity right-parenthesis
left-parenthesis negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
left-parenthesis negative infinity comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
left-bracket negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
left-parenthesis negative infinity comma negative 1 right-parenthesis union left-parenthesis negative 1 comma 1 right-parenthesis union left-parenthesis 1 comma infinity right-parenthesisImage with alt text: left-parenthesis negative infinity comma negative 1 right-parenthesis union left-parenthesis negative 1 comma 1 right-parenthesis union left-parenthesis 1 comma infinity right-parenthesis
left-parenthesis negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
left-parenthesis negative infinity comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
Image with alt text: left-parenthesis negative infinity comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
left-bracket negative 4 comma negative 3 right-parenthesis union left-parenthesis negative 3 comma infinity right-parenthesis
The correct answer is:
(-∞, -4) U (-4, -3) U (-3, ∞)
(-∞, -4) U (-4, -3) U (-3, ∞)
a,b,c or d
d