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
To make the function f(x) = (x-5)^2 + 2 one-to-one, we can restrict the domain.
Let's consider the original function, f(x) = (x-5)^2 + 2.
Notice that the function is a parabola with its vertex at (5,2), opening upwards. This means that for any given y-coordinate, there are two corresponding x-coordinates. Therefore, the function is not one-to-one.
To make the function one-to-one, we can restrict the domain to one side of the vertex. Let's choose the right side of the vertex (x ≥ 5). This means we will only consider x-values equal to or greater than 5.
Now, let's find the inverse function.
1. Start with the original function: f(x) = (x-5)^2 + 2.
2. Replace f(x) with y: y = (x-5)^2 + 2.
3. Swap x and y: x = (y-5)^2 + 2.
4. Solve for y:
x - 2 = (y-5)^2.
√(x - 2) = y - 5.
y = √(x - 2) + 5.
So, the inverse function is f^(-1)(x) = √(x - 2) + 5, where x is limited to values equal to or greater than 5.
To identify a restricted domain that makes the function f(x) = (x-5)^2 + 2 one-to-one, we need to restrict the domain to a specific range of x values where the function does not repeat any y values.
Let's first consider the original function f(x) = (x-5)^2 + 2. This is a quadratic function with a vertex at (5, 2). By observing the graph of this function, we can see that it opens upwards and has a minimum point at the vertex. This means that the range of the function (y-values) starts from the minimum value and increases indefinitely.
To make the function one-to-one, we can choose a restricted domain that excludes values of x where the graph is not increasing or where it has repeated y-values. In this case, we can choose a restricted domain where x is either less than 5 or greater than 5.
Let's consider the restricted domain where x < 5. In this case, we have a subset of the original function f(x). Let's call this function g(x) within this domain.
g(x) = (x-5)^2 + 2 where x < 5
To find the inverse function of g(x), we can swap the x and y variables and solve for y.
x = (y-5)^2 + 2
x - 2 = (y-5)^2
√(x - 2) = y - 5
√(x - 2) + 5 = y
The inverse function of g(x) is given by:
g^(-1)(x) = √(x - 2) + 5 where x < 5
To determine a restricted domain that makes the given function, f(x) = (x - 5)^2 + 2, one-to-one, we need to identify where the function fails to satisfy the horizontal line test.
The horizontal line test states that a function is one-to-one if every horizontal line intersects the graph at most once.
In this case, the function is a parabola that opens upwards, which means it is not one-to-one over its entire domain since there are multiple x-values that produce the same y-value.
To make the function one-to-one, we can restrict the domain. Let's consider restricting it to an interval where the function is strictly increasing or strictly decreasing.
Since the function is a parabola that opens upwards, its vertex (the minimum point) represents the lowest y-value. Thus, the function is strictly increasing on the interval before the vertex and strictly decreasing after the vertex.
To find the vertex, we can rewrite the function in vertex form by completing the square:
f(x) = (x - 5)^2 + 2
= x^2 - 10x + 25 + 2
= x^2 - 10x + 27
The vertex of a quadratic function in the form f(x) = ax^2 + bx + c is given by (-b/2a, f(-b/2a)).
For our function, a = 1 and b = -10. Therefore, the x-coordinate of the vertex is -(-10) / (2 * 1) = 5.
The y-coordinate of the vertex can be found by substituting the x-coordinate back into the function:
f(5) = 5^2 - 10(5) + 27
= 25 - 50 + 27
= 2
So, the vertex of the parabola is (5, 2).
To restrict the domain and make the function one-to-one, we can consider two intervals: x ≤ 5 and x ≥ 5.
1. For x ≤ 5, the function is strictly decreasing.
2. For x ≥ 5, the function is strictly increasing.
Let's choose the interval x ≤ 5 to find the inverse function.
To find the inverse function, let y = f(x) and swap the variables:
x = (y - 5)^2 + 2
Swap x and y:
y = (x - 5)^2 + 2
This is the inverse function of f(x) = (x - 5)^2 + 2 when x ≤ 5.
Therefore, the restricted domain for f(x) to make it one-to-one is x ≤ 5, and the inverse function is g(x) = (x - 5)^2 + 2.