Question

Let y = 2(x – 5)2 – 6.

Part A: Is the given relation a function? Is it one-to-one? Explain completely. If it is not one-to-one, determine a possible restriction on the domain such that the relation is one-to-one.

Part B: Determine y–1. Show all necessary calculations.

Part C: Prove algebraically that y and y–1 are inverse functions.

Answer 1

Not one one

• y=2(x-5)²-6

It's quadratic function as degree is 2

Parabola will be produced if we graph it

and it's parabola is symmetric on both sides of vertex so it will be failed in one line test as a single line can cross 2 points on graph .

Vertex is axis of symmetry

• Here that's x=5 as vertex is at(5,-6)

So

For being one one

• (x-5)>5
• x>10

The domain must contain values greater than 10

As

• f(5) is -6

Any values more than 5 will be on one side of symmetry and then function is oneone

#2

We need inverse of y

Exchange variables

• x=2(y-5)²-6

Solve y

• 2(y-5)²=x+6
• (y-5)²=x+6/2
• y=√(x+6/2)+5

That's inverse

inverse one be g(x) and first one be f(x)

There are two ways .

1. Either find fog(x) and gof(x) if equal value arises then they are inverses
2. Or Put any constant value on both functions, the coordinates should turn like (x,y)=(y,x)

So

Let's check for 0

g(0)

Using calculator

• 6.7

f(6.7)

• 0 nearly

The points are

• (6.7,0) and (0,6.7)

They satisfy (x,y)=(y,x)

Hence they are inverses of each other

Answer 2

Answer:

Given equation:

$y=2(x-5)^2-6$

Vertex form of a quadratic equation:

 $y=a(x-h)^2+k$

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 test
• Algebraic 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):

$\implies f(a) = f(b)$

$\implies 2(a-5)^2-6=2(b-5)^2-6$

$\implies 2(a-5)^2=2(b-5)^2$

$\implies (a-5)^2=(b-5)^2$

$\implies \pm \sqrt{a-5}=\pm \sqrt{b-5}$

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 $\sf y^{-1}$, rearrange the equation to make x the subject:

$\implies y=2(x-5)^2-6$

$\implies y+6=2(x-5)^2$

$\implies \dfrac{y+6}{2}=(x-5)^2$

$\implies \pm\sqrt{\dfrac{y+6}{2}}=x-5$

$\implies x=5\pm\sqrt{\dfrac{y+6}{2}}$

Replace x with $\sf y^{-1}$ and y with x:

$\implies y^{-1}=5\pm\sqrt{\dfrac{x+6}{2}}. \quad x\geq -6$

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))

$\textsf{Let }g(x)=5+\sqrt{\dfrac{x+6}{2}}$

$\textsf{Let }f(x)=2(x-5)^2-6$

$\implies g\left[f(x)\right]=5+\sqrt{\dfrac{\left[2(x-5)^2-6\right]+6}{2}}$

$\implies g\left[f(x)\right]=5+\sqrt{\dfrac{2(x-5)^2}{2}}$

$\implies g\left[f(x)\right]=5+\sqrt{(x-5)^2}$

$\implies g\left[f(x)\right]=5+(x-5)$

$\implies g\left[f(x)\right]=x$

$\implies f\left[g(x)\right]=2\left(5\pm\sqrt{\dfrac{x+6}{2}}}-5\right)^2-6$

$\implies f\left[g(x)\right]=2\left(\pm\sqrt{\dfrac{x+6}{2}}}\right)^2-6$

$\implies f\left[g(x)\right]=2\left(\dfrac{x+6}{2}\right)-6$

$\implies f\left[g(x)\right]=x+6-6$

$\implies f\left[g(x)\right]=x$