Animated Solution for Mathematics - Quadratic Equations: For real $x$, the function $\frac{(x-a)(x-b)}{x-c}$ will assume all real values provided

The rational function is defined as $f(x)=\frac{(x-a)(x-b)}{x-c}$. The problem asserts that this function assumes all real values, meaning the range of $f(x)$ is $(-\infty ,\infty )$.

To analyze this, we set $y=\frac{(x-a)(x-b)}{x-c}$. For $y$ to span all real numbers, there must exist at least one real $x$ for every chosen $y$.

We begin by cross-multiplying the denominator:

Expanding both sides yields:

Rearranging the terms into a standard quadratic equation in terms of $x$, we obtain:

Since $x$ must be a real number, the discriminant $D_x$ of this quadratic equation must be non-negative ($D_x\ge 0$). Calculating the discriminant:

Expanding the squared term $(a+b+y{)}^2$ as $(a+b{)}^2+y^2+2y(a+b)$, the inequality becomes:

Grouping the terms by powers of $y$, we arrive at a new quadratic inequality in $y$:

The problem states that $y$ takes all real values. Therefore, the inequality $y^2+2y(a+b-2c)+(a-b{)}^2\ge 0$ must hold for all $y\in \mathbb{R}$.

For a quadratic $A{y}^2+By+C$ to be non-negative for all $y$, its own discriminant $D_y$ must be less than or equal to zero ($D_y\le 0$). We calculate $D_y$:

Dividing the inequality by 4, we simplify the expression to:

Applying the difference of squares identity $X^2-Y^2=(X-Y)(X+Y)$, where $X=(a+b-2c)$ and $Y=(a-b)$:

Simplifying the terms inside the brackets:

Dividing by 4, we reach the final condition:

This result implies that $c$ must lie between $a$ and $b$ (inclusive of the endpoints if the function remains defined). This is the necessary and sufficient condition for the rational function to assume all real values.