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

Jul 11, 2015

$\frac{d}{\mathrm{dx}} f \left(x\right) = 4 x - 4 = 4 \left(x - 1\right) \ge 0$ when $x \ge 1$ and is only 0 when $x = 1$.

So $f \left(x\right)$ is strictly monotonically increasing for $x \in \left[1 , \infty\right)$ so is 1-1

#### Explanation:

This is a parabola with vertex at $x = 1$ so is strictly monotonic for $x \in \left[1 , \infty\right)$ so one to one.

Jul 11, 2015

If ${x}_{1} , {x}_{2} \in \left[1 , \infty\right)$ show that $f \left({x}_{1}\right) = f \left({x}_{2}\right) \implies {x}_{1} = {x}_{2}$

by directly solving. So $f \left(x\right)$ is 1-1 on $\left[1 , \infty\right)$

#### Explanation:

$f \left(x\right) = 2 {x}^{2} - 4 x - 1 = 2 {\left(x - 1\right)}^{2} - 3$

So if $f \left({x}_{1}\right) = f \left({x}_{2}\right)$ then

$2 {\left({x}_{1} - 1\right)}^{2} - 3 = 2 {\left({x}_{2} - 1\right)}^{2} - 3$

Add $3$ to both sides to get:

$2 {\left({x}_{1} - 1\right)}^{2} = 2 {\left({x}_{2} - 1\right)}^{2}$

Divide both sides by $2$ to get:

${\left({x}_{1} - 1\right)}^{2} = {\left({x}_{2} - 1\right)}^{2}$

So:

$\left\mid {x}_{1} - 1 \right\mid = \left\mid {x}_{2} - 1 \right\mid$

Given ${x}_{1} \ge 1$ and ${x}_{2} \ge 1$ then

${x}_{1} - 1 \ge 0$ so $\left\mid {x}_{1} - 1 \right\mid = {x}_{1} - 1$

${x}_{2} - 1 \ge 0$ so $\left\mid {x}_{2} - 1 \right\mid = {x}_{2} - 1$

So

${x}_{1} - 1 = \left\mid {x}_{1} - 1 \right\mid = \left\mid {x}_{2} - 1 \right\mid = {x}_{2} - 1$

Add $1$ to both ends to get:

${x}_{1} = {x}_{2}$

So $f \left({x}_{1}\right) = f \left({x}_{2}\right) \implies {x}_{1} = {x}_{2}$ for all ${x}_{1} , {x}_{2} \in \left[1 , \infty\right)$

Jul 11, 2015

This can also be determined graphically.

#### Explanation:

The graph of $f \left(x\right) = 2 {x}^{2} - 4 x - 1$ is a parabola with vertex: $\left(1 , - 3\right)$

(Use the vertex formula or put the expression in vertex form: $f \left(x\right) = 2 {\left(x - 1\right)}^{2} - 3$.)

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 $g \left(x\right) = 2 {x}^{2} - 4 x - 1$ (domain $\left(- \infty , \infty\right)$) 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 $f \left(x\right)$ 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.