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

3 Answers
Jul 11, 2015

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.

Jul 11, 2015

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)

Jul 11, 2015

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.