Animated Solution for Mathematics - Functions: The function $f:\mathbb{R}\to \mathbb{R},f(x)=\frac{x^2+2x-15}{x^2-4x+9},x\in \mathbb{R}$ is

Welcome, fellow explorers of mathematics! Today, we are going to dissect a rational function. It might look like a simple fraction of two quadratics, but within its structure lies a beautiful story of mapping, constraints, and the fundamental nature of functions.

We are investigating the function $f:\mathbb{R}\to \mathbb{R}$ defined by:

Our goal is to determine if it is one-one, onto, or neither.

To be 'one-one' (or injective), every unique input $x$ must map to a unique output $y$. If we can find two different inputs that land on the same output, the function is 'many-one' and fails the one-one test.

A quick way to test for many-one behavior is to see if the function hits the same value twice. Let's check the roots by setting $f(x)=0$. This forces the numerator to be zero:

Factoring this quadratic is a breeze: $(x+5)(x-3)=0$. This gives us two distinct roots: $x=-5$ and $x=3$.

Think about what this means. Both $x=-5$ and $x=3$ map to the output $y=0$. Because two distinct elements in the domain share the same image, the function is definitively many-one.

Now, we turn our attention to the 'onto' (or surjective) property. A function is onto if its range is equal to its codomain. Here, the codomain is given as $\mathbb{R}$ (all real numbers).

If we find that the range is anything less than the entire set of real numbers, the function is not onto. To find the range, we set $y=f(x)$ and treat $y$ as a parameter:

Cross-multiplying gives us:

Now, let's gather our terms to form a quadratic equation in $x$:

This is the pivotal moment. For $x$ to be a real number, the discriminant $D$ of this quadratic equation must be greater than or equal to zero ($D\ge 0$). This is the 'gatekeeper' condition that defines our range.

Using the formula $D=B^2-4AC$, where $A=(y-1)$, $B=-2(2y+1)$, and $C=(9y+15)$, we write:

Dividing by 4 makes the numbers much friendlier:

Expanding the terms:

To make this inequality easier to handle, multiply by $-1$ (remembering to flip the inequality sign!):

Factoring this quadratic gives us $(5y-8)(y+2)\le 0$. The critical points are $y=-2$ and $y=\frac{8}{5}$ (or $1.6$). Using the wavy curve method, we see that $y$ must lie in the interval $[-2,1.6]$.

We have discovered that the range of our function is $[-2,1.6]$. Compare this to the codomain, which is the entire set of real numbers $\mathbb{R}$.

Since the range is clearly not equal to the codomain, the function is not onto. We have proven that the function is neither one-one nor onto.