Question

Show that the function f:R--(-5, infinite) defined by f(x)=9x² +6x-5 is one one and onto function​

Answer 1

Answer:

The given function is one one and onto.

Step-by-step explanation:

Given the function $f:\mathbb{R}\to(-5,\infty)$ defined by,

$f(x)=9x^2+6x-5$

A function is one one if images of different numbers are different.

If

$f(x)=f(y)\\\implies 9x^2+6x-5=9y^2+6y-5\\\implies 9x^2+6x=9y^2+6y\\\implies 3x(3x+2)=3y(3y+2)\\\implies x(3x+2)=y(3y+2)$

This happens if and only if $x=y$.

Now to prove that it is onto we take and element in the codomain and prove that it has a preimage.

Let $y\in \mathbb{R}$, then we shall check that,

$9x^2 +6x-5=y$ has a solution or not

$9x^2+6x-5-y=0$

Since

$6^2-4\times9\times(-5-y)=36+36(5+y)$ is non negative if $y\geq -5$. So solution exists for all elements in $(-5,\infty)$. So the function is onto.