Answer:
Given equation:
[tex]y=2(x-5)^2-6[/tex]
Vertex form of a quadratic equation:
[tex]y=a(x-h)^2+k[/tex]
where (h, k) is the vertex
Therefore, the given equation is a quadratic equation in vertex form with a vertex at (5, -6) and a range of [-6, ∞)
Part A
One-to-one functions have a unique x-value for every y-value.
As given equation is a quadratic equation in vertex form it is therefore not one-to-one function.
There are two ways to test if a function is one-to-one:
Horizontal line testAlgebraic Testing
Horizontal line test
If a horizontal line intersects a function's graph more than once, then the function is not one-to-one.
As the given function is quadratic, its graph is a parabola. Therefore, there will be 2 values of x for each value of y (with the exception of the vertex) and so the function fails the horizontal line test.
Algebraic Testing
The function is one-to-one if a = b for every f(a) = f(b):
[tex]\implies f(a) = f(b)[/tex]
[tex]\implies 2(a-5)^2-6=2(b-5)^2-6[/tex]
[tex]\implies 2(a-5)^2=2(b-5)^2[/tex]
[tex]\implies (a-5)^2=(b-5)^2[/tex]
[tex]\implies \pm \sqrt{a-5}=\pm \sqrt{b-5}[/tex]
Therefore, it is not a one-to-one function.
To make the given relation one-to-one we need to restrict the domain.
The x-value of the vertex of the function is x = 5, therefore its axis of symmetry is x = 5.
So to make the given relation one-to-one, the domain should be:
(-∞, 5] or [5, ∞)
Part B
To find the inverse [tex]\sf y^{-1}[/tex], rearrange the equation to make x the subject:
[tex]\implies y=2(x-5)^2-6[/tex]
[tex]\implies y+6=2(x-5)^2[/tex]
[tex]\implies \dfrac{y+6}{2}=(x-5)^2[/tex]
[tex]\implies \pm\sqrt{\dfrac{y+6}{2}}=x-5[/tex]
[tex]\implies x=5\pm\sqrt{\dfrac{y+6}{2}}[/tex]
Replace x with [tex]\sf y^{-1}[/tex] and y with x:
[tex]\implies y^{-1}=5\pm\sqrt{\dfrac{x+6}{2}}. \quad x\geq -6[/tex]
Part C
A function g(x) is said to be an inverse of a function f(x) if g(f(x))=x=f(g(x))
[tex]\textsf{Let }g(x)=5+\sqrt{\dfrac{x+6}{2}}[/tex]
[tex]\textsf{Let }f(x)=2(x-5)^2-6[/tex]
[tex]\implies g\left[f(x)\right]=5+\sqrt{\dfrac{\left[2(x-5)^2-6\right]+6}{2}}[/tex]
[tex]\implies g\left[f(x)\right]=5+\sqrt{\dfrac{2(x-5)^2}{2}}[/tex]
[tex]\implies g\left[f(x)\right]=5+\sqrt{(x-5)^2}[/tex]
[tex]\implies g\left[f(x)\right]=5+(x-5)[/tex]
[tex]\implies g\left[f(x)\right]=x[/tex]
[tex]\implies f\left[g(x)\right]=2\left(5\pm\sqrt{\dfrac{x+6}{2}}}-5\right)^2-6[/tex]
[tex]\implies f\left[g(x)\right]=2\left(\pm\sqrt{\dfrac{x+6}{2}}}\right)^2-6[/tex]
[tex]\implies f\left[g(x)\right]=2\left(\dfrac{x+6}{2}\right)-6[/tex]
[tex]\implies f\left[g(x)\right]=x+6-6[/tex]
[tex]\implies f\left[g(x)\right]=x[/tex]
Part A: The given relation is a function but not one-to-one due to the parabolic shape. Restricting the domain to x ≥ 5 makes it one-to-one. Part B: The inverse function y–1 is x = 5 ± √((y + 6)/2). Part C: The algebraic proof shows that y and y–1 are inverse functions as their compositions result in the identity function y(y–1) = 25.
Part A: The given relation is a function because for each value of "x," there is only one corresponding value of "y" based on the equation. However, to determine if it is one-to-one, we need to check if each unique "x" value corresponds to a unique "y" value.
The relation is not one-to-one as it stands. This is because the term "2(x – 5)²" creates a parabolic shape, and for each value of "x," there are two possible "y" values due to the squared term. For example, if we take x = 0 and x = 10, both will yield the same "y" value of -6.
To make the relation one-to-one, we can restrict the domain. If we restrict x to values greater than or equal to 5, the relation will become one-to-one. This is because the graph will be limited to the right side of the vertex of the parabola, ensuring that each "x" value has a unique corresponding "y" value.
Part B: To determine y–1 (the inverse of the function), we need to solve the given equation for "x." Let's start by reversing the steps in the equation:
y = 2(x – 5)² – 6
y + 6 = 2(x – 5)²
(y + 6)/2 = (x – 5)²
Now, take the square root of both sides:
±√((y + 6)/2) = x – 5
x = 5 ± √((y + 6)/2)
So, the inverse function is x = 5 ± √((y + 6)/2).
Part C: To prove algebraically that y and y–1 are inverse functions, we need to show that their compositions result in the identity function.
Starting with the original equation for y:
y = 2(x – 5)² – 6
Replace "y" with "x" and solve for "x":
x = 2(y – 5)² – 6
x + 6 = 2(y – 5)²
(x + 6)/2 = (y – 5)²
±√((x + 6)/2) = y – 5
y = 5 ± √((x + 6)/2)
This is the same expression we derived for the inverse function y–1.
Now, if we substitute y–1 into y and simplify:
y(y–1) = (5 ± √((x + 6)/2))(5 ± √((y + 6)/2))
Using the difference of squares formula, we can simplify further:
y(y–1) = (25 - ((x + 6)/2))
Simplifying again:
y(y–1) = 25 - (x + 6)/2
Notice that the term (x + 6)/2 is canceled out by the algebraic manipulation, resulting in 25. Therefore, y(y–1) = 25, which is the identity function.
This proves that y and y–1 are indeed inverse functions.
To know more about function:
https://brainly.com/question/31062578
#SPJ3
