The function lower f left parenthesis x right parenthesis equals x superscript 2 baseline plus 2 is not one-to-one. Determine a restricted domain that makes it one-to-one, and find the inverse function.
To determine a restricted domain that makes the function one-to-one, we need to find a range of values for x where each input has a unique output.
Since the function is a quadratic function, we know that the graph is a parabola that opens upward. This means that the function will not be one-to-one if the parabola intersects the x-axis at more than one point.
To find the restricted domain, we need to find where the parabola intersects the x-axis. Setting the function equal to zero:
x^2 + 2 = 0
We can see that the polynomial can never be equal to zero since x^2 is always non-negative. Therefore, the parabola does not intersect the x-axis, and the function is one-to-one for all real values of x.
To find the inverse function, we can switch the roles of x and y and solve for y.
x = y^2 + 2
Rearranging the equation:
y^2 = x - 2
Taking the square root of both sides:
y = ±√(x - 2)
Since we are looking for the inverse, we only want a single value for y. Hence, we take the positive square root:
y = √(x - 2)
Therefore, the inverse function of f(x) = x^2 + 2 is g(x) = √(x - 2).
answer choices
a) restricted domain: x less-than-or-equal-to 2; lower f superscript negative 1 baseline left parenthesis x right parenthesis equals negative Start Root x minus 2 End Root
restricted domain: Image with alt text: x less-than-or-equal-to 2 ; Image with alt text: lower f superscript negative 1 baseline left parenthesis x right parenthesis equals negative Start Root x minus 2 End Root
b) restricted domain: x less-than-or-equal-to 0; lower f superscript negative 1 baseline left parenthesis x right parenthesis equals Start Root x plus 2 End Root
restricted domain: Image with alt text: x less-than-or-equal-to 0 ; Image with alt text: lower f superscript negative 1 baseline left parenthesis x right parenthesis equals Start Root x plus 2 End Root
c) restricted domain: x greater-than-or-equal-to 0; lower f superscript negative 1 baseline left parenthesis x right parenthesis equals Start Root x minus 2 End Root
restricted domain: Image with alt text: x greater-than-or-equal-to 0 ; Image with alt text: lower f superscript negative 1 baseline left parenthesis x right parenthesis equals Start Root x minus 2 End Root
d) restricted domain: x greater-than-or-equal-to 0; lower f superscript negative 1 baseline left parenthesis x right parenthesis equals negative Start Root x minus 2 End Root
The correct answer choice is:
c) restricted domain: x greater-than-or-equal-to 0; the inverse function is: lower f superscript negative 1 baseline left parenthesis x right parenthesis equals Start Root x minus 2 End Root
Determine the domain of the function left parenthesislower f circle lower g right parenthesis left parenthesis x right parenthesis where lower f left parenthesis x right parenthesis equals Start Fraction 3 over x plus 2 End Fraction and lower g left parenthesis x right parenthesis equals Start Fraction 5 over x End Fraction.
To find the domain of the composition of functions f(g(x)), we need to consider the restrictions on both functions f(x) and g(x).
For the function f(x) = 3/(x + 2), the only restriction is that the denominator (x + 2) cannot be zero. So, x + 2 ≠ 0.
Solving this inequality:
x ≠ -2
Therefore, the domain of f(x) is all real numbers except -2.
For the function g(x) = 5/x, the restriction is that the denominator (x) cannot be zero. So, x ≠ 0.
Therefore, the domain of g(x) is also all real numbers except 0.
To find the domain of the composition f(g(x)), we need to consider the restrictions on both f(x) and g(x). In this case, the denominators of both f(x) and g(x) cannot be zero, so x + 2 ≠ 0 and x ≠ 0.
Combining these inequalities:
x + 2 ≠ 0 and x ≠ 0
We can combine these conditions by taking their intersection:
x ≠ -2 and x ≠ 0
So, the domain of the composition of functions f(g(x)) is all real numbers except -2 and 0.
answer choices
a) left parenthesis negative infinity comma negative 2 right parenthesis union left parenthesis negative 2 comma infinity right parenthesisImage with alt text: left parenthesis negative infinity comma negative 2 right parenthesis union left parenthesis negative 2 comma infinity right parenthesis
b) left parenthesis negative infinity comma negative Start Fraction 2 over 5 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 5 comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma negative Start Fraction 2 over 5 End Fraction right parenthesis union left parenthesis negative Start Fraction 2 over 5 comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
c) left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
Image with alt text: left parenthesis negative infinity comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
d) left parenthesis negative infinity comma negative Start Fraction 5 over 2 End Fraction right parenthesis union left parenthesis negative Start Fraction 5 over 2 comma 0 right parenthesis union left parenthesis 0 comma infinity right parenthesis
The correct answer choice is:
b) (negative infinity, -2) union (-2, infinity)
wrong
Apologies for the confusion. Let's reconsider the domain of the function.
The function f(g(x)) is given by f(g(x)) = f(5/x) = 3/(5/x + 2).
To determine the domain of the function f(g(x)), we need to consider the restrictions on both f(x) and g(x).
For the function f(x) = 3/(x + 2), the denominator (x + 2) cannot be zero. So, x + 2 ≠ 0.
Solving this inequality:
x ≠ -2
Therefore, the domain of f(x) is all real numbers except -2.
For the function g(x) = 5/x, the denominator (x) cannot be zero. So, x ≠ 0.
Therefore, the domain of g(x) is also all real numbers except 0.
To find the domain of the composition f(g(x)), we need to consider the restrictions on both f(x) and g(x). In this case, the denominators of both f(x) and g(x) cannot be zero, so x + 2 ≠ 0 and x ≠ 0.
Combining these inequalities:
x + 2 ≠ 0 and x ≠ 0
We can combine these conditions by taking their intersection:
x ≠ -2 and x ≠ 0
So, the domain of the composition of functions f(g(x)) is all real numbers except -2 and 0.
The correct answer choice is:
a) (-∞, -2) ∪ (-2, ∞)