Animated Solution for Mathematics - Basic Mathematics: Let $y=\frac{(x+1)(x-3)}{(x-2)}$. Find all the real values of $x$ for which $y$ takes real values.

Welcome, future engineer. Today, we are not just solving an inequality; we are mapping the landscape of a function. When you look at $y=\frac{(x+1)(x-3)}{(x-2)}$, don't just see a jumble of variables.

See a story of balance, boundaries, and critical transitions. Our goal is to find where this function lives in the 'real' world—specifically, where it remains non-negative.

We set our inequality as:

Before we touch a pen to paper, let's analyze our components. We have a numerator, $(x+1)(x-3)$, and a denominator, $(x-2)$.

The numerator tells us where the function hits the x-axis—these are our roots, where $y=0$. The denominator, however, tells us where the function breaks.

If $x=2$, we are dividing by zero, and the function ceases to exist. This is our forbidden zone.

Think of these points as the 'hinges' of our function. By setting each factor to zero, we find the points where the sign of the entire expression is forced to change:

1. $x+1=0\Rightarrow x=-1$ 2. $x-3=0\Rightarrow x=3$ 3. $x-2=0\Rightarrow x=2$

These three points, $-1$, $2$, and $3$, divide the entire number line into four distinct regions. Our job is to determine which of these regions satisfy our condition of being greater than or equal to zero.

Now, let's test the behavior. If we pick a value to the right of our largest root, say $x=4$, we calculate:

Since $2.5>0$, we know the function is positive in the interval $(3,\infty )$. Because every factor in our expression is raised to an odd power (the power of 1), the function must flip its sign every time we cross a critical point.

It’s like a wave passing through these points:

For $x>3$, the expression is positive ($+$). Crossing $x=3$, it becomes negative ($-$) in the interval $(2,3)$. Crossing $x=2$, it becomes positive ($+$) in the interval $(-1,2)$. Crossing $x=-1$, it becomes negative ($-$) in the interval $(-\infty ,-1)$.

We are looking for regions where the expression is $\ge 0$. Looking at our wave, we see two regions that satisfy this: the interval between $-1$ and $2$, and the interval from $3$ onwards.

But wait—we must respect the boundaries! At $x=-1$ and $x=3$, the expression is exactly zero, which is allowed.

We use square brackets $[-1]$ and $[3]$ to show these points are included. At $x=2$, the expression is undefined, so we must use a parenthesis $(2)$ to show that $2$ is strictly excluded.

Thus, our final solution set is:

This isn't just an algebraic result; it is the set of all real numbers that allow this function to exist in the positive realm. You have successfully navigated the critical points and mapped the behavior of the function.