Animated Solution for Mathematics - Functions: If $y=f(x)=\frac{x+2}{x-1}$ then

Welcome, future engineer! Today, we are going to dissect a function that might look simple at first glance, but holds within it the core principles of algebra and calculus.

We are looking at the function:

This is a classic rational function, and understanding its behavior is a rite of passage for every JEE aspirant.

Defining the Beast

First, let us address the definition. A rational function is defined as $R(x)=\frac{P(x)}{Q(x)}$, where both $P(x)$ and $Q(x)$ are polynomials.

In our case, $P(x)=x+2$ and $Q(x)=x-1$. Both are linear polynomials.

Therefore, by definition, our function is indeed a rational function. This is the foundation upon which we build our analysis.

The Forbidden Zone

Now, let us talk about the domain. Every function has its boundaries, and for rational functions, the boundary is the denominator.

We ask ourselves: where does the denominator vanish? Setting $x-1=0$, we find $x=1$.

At this point, the function is undefined. Geometrically, this creates a vertical asymptote. If you were to graph this, you would see the curve approaching infinity as it gets closer to $x=1$.

This immediately tells us that $f(1)$ does not exist. The function is undefined at $x=1$.

The Mirror Image

Next, let us explore the inverse. The question asks if $x=f(y)$. This is a sophisticated way of asking if the function is its own inverse.

To find out, we start with:

Our goal is to isolate $x$. We cross-multiply to get $y(x-1)=x+2$. Expanding this, we have $xy-y=x+2$.

Now, we gather all terms containing $x$ on one side: $xy-x=y+2$. Factoring out $x$, we get $x(y-1)=y+2$.

Finally, dividing by $(y-1)$, we obtain:

Look closely! This is the exact same algebraic structure as our original function $f(x)$. This means $f(x)$ is an involution—it is its own inverse. This is a beautiful, symmetric result.

The Slope of the Curve

Finally, we investigate the monotonicity. Does the function increase or decrease? To answer this, we need the derivative $f^{\prime }(x)$.

Using the quotient rule, $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$, where $u=x+2$ and $v=x-1$, we calculate:

Simplifying the numerator, we get $x-1-x-2=-3$. Thus:

Since $(x-1{)}^2$ is always positive for $x eq 1$, the derivative $f^{\prime }(x)$ is always negative. A negative derivative means the function is strictly decreasing everywhere in its domain.

Therefore, the claim that it increases for $x<1$ is incorrect.

We have successfully navigated the properties of this function. We identified it as a rational function, understood the vertical asymptote at $x=1$, discovered the beautiful self-inverse property, and used the derivative to prove it is strictly decreasing.

Keep practicing these steps, and you will find that even the most complex functions start to reveal their secrets to you. You are doing great!