The function f(x)=(x-5)^2+2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the inverse function.

a. restricted domain: x>or=5; f^-1(x)=5-sqrt(x-2)

b. restricted domain: x>or=5; f^-1(x)=5+sqrt(x-2)

c. restricted domain: x<or=5; f^-1(x)=5+sqrt(x+2)

d. restricted domain: x<or=-5; f^-1(x)=5+sqrt(x-2)

I really need help if someone could please explain this to me, I would greatly appreciate it. Thank you!

the axis of symmetry is the line x=5

So, any domain on one side of that line will be one-to-one.

Now, for the inverse, swap variables and solve for f^-1

x = (f-5)^2+2
x-2 = (f-5)^2
f-5 = ±√(x-2)
f = 5±√(x-2)

Now, we need one branch only, but √(x-2) is only defined for x>=2

On the original parabola, the right branch is x>=5. So, only A or B is a candidate, and so B is the choice.

Sure, I'd be happy to help!

To determine a restricted domain that makes the function one-to-one, we need to find a region where the function passes the horizontal line test. This means that any horizontal line should intersect the graph of the function at most once.

To find the restricted domain, we can look at the behavior of the function's graph. The given function is a quadratic function in the form f(x) = (x - h)^2 + k, where (h, k) is the vertex of the parabola. In this case, the vertex is (5, 2).

Since the parabola opens upwards (because the coefficient of the x^2 term is positive), the graph will be symmetric with respect to the vertical line x = 5. This means that any horizontal line that intersects the graph will intersect it in two places, except for the horizontal line passing through the vertex.

In order to make the function one-to-one, we need to exclude the vertex from the domain. Since the vertex is at x = 5, we can choose a restricted domain of either x ≥ 5 or x ≤ 5.

Given this restricted domain, we can find the inverse function by switching the roles of x and y and solving for y.

In this case, the correct answer is option a.
The restricted domain is x ≥ 5. The inverse function is f^−1(x) = 5 − sqrt(x − 2).

I hope this clarifies things! Let me know if you have any other questions.

To determine a restricted domain that makes the function f(x) = (x-5)^2 + 2 one-to-one, we need to identify a section of the function where no two different inputs x yield the same output f(x). In other words, we need to find a range of x values where the function is strictly increasing or decreasing.

The given function is a quadratic function in the form of f(x) = (x-5)^2 + 2. It opens upward (since the coefficient of x^2 is positive) and the vertex is at (5, 2). The vertex represents the minimum value of the function.

To restrict the domain and make the function one-to-one, we can choose either all x values greater than or equal to 5 (x ≥ 5) or all x values less than or equal to 5 (x ≤ 5). Let's consider each option:

a) restricted domain: x ≥ 5
In this case, the quadratic function will be strictly increasing. The smallest value the function can take is f(5) = 2, and from there on, it increases as x increases. Thus, this restricted domain will make the function one-to-one.

The inverse function is found by switching the roles of x and y and solving for y. Let's call the inverse function f^-1(x):
x = (y-5)^2 + 2
x - 2 = (y - 5)^2
√(x - 2) = y - 5
y = 5 + √(x - 2)

Therefore, the correct answer is: a) restricted domain: x ≥ 5; f^-1(x) = 5 + √(x - 2).

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