Question

How to determine whether f(x)= 2x²-4x -1 is one to one function given that x is more than or equal to 1?

Answer 1

`d/(dx) f(x) = 4x-4 = 4(x-1) >= 0` when `x >= 1` and is only 0 when `x=1`.

So `f(x)` is strictly monotonically increasing for `x in [1, oo)` so is 1-1

Explanation:

This is a parabola with vertex at `x=1` so is strictly monotonic for `x in [1, oo)` so one to one.

Answer 2

If `x_1, x_2 in [1, oo)` show that `f(x_1) = f(x_2) => x_1 = x_2`

by directly solving. So `f(x)` is 1-1 on `[1, oo)`

Explanation:

`f(x) = 2x^2-4x-1 = 2(x-1)^2-3`

So if `f(x_1) = f(x_2)` then

`2(x_1-1)^2-3 = 2(x_2-1)^2-3`

Add `3` to both sides to get:

`2(x_1-1)^2 = 2(x_2-1)^2`

Divide both sides by `2` to get:

`(x_1-1)^2 = (x_2-1)^2`

So:

`abs(x_1-1) = abs(x_2-1)`

Given `x_1 >= 1` and `x_2 >= 1` then

`x_1 - 1 >= 0` so `abs(x_1 - 1) = x_1 - 1`

`x_2 - 1 >= 0` so `abs(x_2 - 1) = x_2 - 1`

So

`x_1 - 1 = abs(x_1 - 1) = abs(x_2 - 1) = x_2 - 1`

Add `1` to both ends to get:

`x_1 = x_2`

So `f(x_1) = f(x_2) => x_1 = x_2` for all `x_1, x_2 in [1, oo)`

Answer 3

This can also be determined graphically.

Explanation:

The graph of `f(x) = 2x^2-4x-1` is a parabola with vertex: `(1, -3)`

(Use the vertex formula or put the expression in vertex form: `f(x) = 2(x-1)^2 -3`.)

So with no restriction on the domain, the graph looks like:

graph{2x^2-4x-1 [-6.97, 10.81, -5.69, 3.2]}

With its natural domain for the expression, the function `g(x) = 2x^2-4x-1` (domain `(-oo, oo)`) fails the horizontal line test. It is not one-to-one.

When we restrict the domain to `x > = 1`, we eliminate the "left half" of the parabola,

The graph of `f(x)` looks like:

graph{y = (2x^2-4x-1)*sqrt(x-1)/sqrt(x-1) [-4.32, 8.17, -4.367, 1.88]}

This function passes the horizontal line test. It is one-to-one.