Animated Solution for Mathematics - Functions: Let $f:\mathbb{R}\to \mathbb{R}$ be defined as $f(x)=x-1$ and $g:\mathbb{R}-\{1,-1\}\to \mathbb{R}$ be defined as $g(x)=\frac{x^2}{x^2-1}$. Then the function $fog$ is :

Welcome, future engineer! Today, we are going to unravel the mystery of composite functions. Imagine you have two machines in a factory: machine $f$ and machine $g$.

When you feed an input $x$ into $g$, it transforms it into $g(x)$. Then, you take that result and feed it into $f$. This is the essence of $f(g(x))$.

We are given the components: $f(x)=x-1$ $g(x)=\frac{x^2}{x^2-1}$

Notice the domain restriction immediately: $x$ cannot be $1$ or $-1$ because that would make the denominator zero. This is the first rule of the game: always respect the domain!

Now, let us define our composite function $h(x)=f(g(x))$. This means we take the entire expression for $g(x)$ and place it into the $x$ slot of $f$.

So, $h(x)=g(x)-1$. Substituting the expression for $g(x)$, we get:

To simplify this, we need a common denominator, which is $x^2-1$. The expression becomes:

Watch closely as the $x^2$ terms vanish—it is like magic! We are left with:

Is it One-One?

To determine if a function is one-one, we ask: does every unique input produce a unique output? Let us test the symmetry.

If we replace $x$ with $-x$, we get:

Since $h(-x)=h(x)$, the function is even. An even function is symmetric about the $y$-axis, which is a dead giveaway that it is many-one.

If you draw a horizontal line at $y=-2$, it will hit the graph at two distinct points. Therefore, it fails the horizontal line test. It is not one-one.

Is it Onto?

Finally, we must determine if the function is onto. This requires us to find the range. We set $y=\frac{1}{x^2-1}$ and solve for $x^2$.

Cross-multiplying gives $y(x^2-1)=1$, which leads to $x^2-1=\frac{1}{y}$. This simplifies to:

Since $x$ is a real number, $x^2$ must be greater than or equal to zero. This forces the inequality:

Using the wavy curve method, we find that $y$ must be in the interval $(-\infty ,-1]\cup (0,\infty )$. This is our range.

The codomain given is $\mathbb{R}$, but our range is missing the interval $(-1,0]$. Because the range does not equal the codomain, the function is into, not onto.

We have successfully dissected the function! It is neither one-one nor onto. Keep practicing this analytical approach, and you will master the art of functions.