A function f : IR → IR, where IR is the set of real numbers, is defined by f(x) = (αx² + 6x - 8) / (

Q: Wait, why do we set the discriminant $D_x \geq 0$?

Great question! We set $y = f(x)$ and rearranged it into a quadratic in $x$. For $x$ to be a real number (which is required for the function to be onto), the quadratic equation must have real roots. The discriminant $D$ tells us exactly when a quadratic has real roots. If $D < 0$, the roots are imaginary, meaning there is no $x$ for that specific $y$, which would violate the 'onto' condition.

Q: Why does the quadratic in $y$ have to be $\geq 0$ for all $y$?

This is the core of the problem. We found that for any $y$, there must exist an $x$. This implies that the condition $D_x \geq 0$ must be true for every single real value of $y$. If we have a quadratic expression in $y$ that must be non-negative for all $y$, it implies the parabola must never dip below the $y$-axis. This forces the leading coefficient to be positive and the discriminant of this $y$-quadratic to be non-positive.

Q: Why is the function not one-to-one at $\alpha = 3$?

Injectivity (one-to-one) means that every output $y$ comes from exactly one input $x$. If we set $\alpha = 3$ and look at the equation $f(x) = 0$, we are essentially asking: 'Which $x$ values map to zero?' If we find two distinct $x$ values that give the same output (zero), the function fails the horizontal line test. Since the discriminant of the numerator was positive, we found two distinct roots, proving it is not one-to-one.

Question

A function $f : IR 	o IR$, where $IR$ is the set of real numbers, is defined by $f(x) = \frac{\alpha x^2 + 6x - 8}{\alpha + 6x - 8x^2}$. Find the interval of values of $\alpha$ for which $f$ is onto. Is the function one-to-one for $\alpha = 3$? Justify your answer.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a vast, infinite landscape representing the codomain of our function, the set of all real numbers $\mathbb{R}$. Our goal is to ensure that our function

$$f(x) = \frac{\alpha x^2 + 6x - 8}{\alpha + 6x - 8x^2}$$

is 'onto'. This means that every single point in this infinite landscape must be reached by at least one input $x$. If even one point is left untouched, the function fails to be onto, which is the essence of the surjective property.

### The Algebraic Bridge

To tackle this, we must build a bridge between the output $y$ and the input $x$. We start by setting:

$$y = \frac{\alpha x^2 + 6x - 8}{\alpha + 6x - 8x^2}$$

By cross-multiplying, we get $y(\alpha + 6x - 8x^2) = \alpha x^2 + 6x - 8$. If we gather all terms on one side, we transform this into a quadratic equation in $x$:

$$(8y + \alpha)x^2 + (6 - 6y)x - (8 + \alpha y) = 0$$

This is the heart of the problem. For the function to be onto, for every $y \in \mathbb{R}$, there must exist at least one real $x$. This is only possible if the quadratic equation we just derived has real roots for every $y$.

### The Discriminant Dance

How do we ensure a quadratic has real roots? We invoke the discriminant, $D_x = b^2 - 4ac$. For real roots, we require $D_x \geq 0$.

Substituting our coefficients, we get:

$$D_x = [6(1 - y)]^2 - 4(8y + \alpha)(-(8 + \alpha y)) \geq 0$$

Expanding this, we get $36(1 - y)^2 + 4(8y + \alpha)(8 + \alpha y) \geq 0$. Dividing by $4$ simplifies our life:

$$9(1 - 2y + y^2) + (64y + 8\alpha y^2 + 8\alpha + \alpha^2 y) \geq 0$$

Combining like terms, we arrive at a new quadratic inequality in $y$:

$$(9 + 8\alpha)y^2 + (\alpha^2 + 46)y + (9 + 8\alpha) \geq 0$$

### The Global Inequality

Here is the crucial realization: this inequality must hold for **all** $y \in \mathbb{R}$. For a quadratic in $y$ to be non-negative for all $y$, it must be an upward-opening parabola that never dips below the $y$-axis.

This gives us two conditions:
1. The leading coefficient must be positive: $9 + 8\alpha > 0$, which means $\alpha > -\frac{9}{8}$.
2. The discriminant of this $y$-quadratic must be less than or equal to zero, $D_y \leq 0$.

Calculating $D_y$:

$$D_y = (\alpha^2 + 46)^2 - 4(9 + 8\alpha)^2 \leq 0$$

Using the difference of squares, we factor this into $(\alpha^2 - 16\alpha + 28)(\alpha^2 + 16\alpha + 64) \leq 0$. This simplifies to:

$$(\alpha - 2)(\alpha - 14)(\alpha + 8)^2 \leq 0$$

Since $(\alpha + 8)^2$ is always non-negative, we focus on $(\alpha - 2)(\alpha - 14) \leq 0$, giving us the interval **$\alpha \in [2, 14]$**.

### The Final Verdict on Injectivity

Finally, we test $\alpha = 3$. The function becomes $f(x) = \frac{3x^2 + 6x - 8}{3 + 6x - 8x^2}$.

To check if it is one-to-one, we check if $f(x) = 0$ has multiple roots. Setting the numerator to zero, $3x^2 + 6x - 8 = 0$, we find the discriminant:

$$D = 36 - 4(3)(-8) = 132$$

Since $D > 0$, there are two distinct real roots. Because two different inputs map to the same output (zero), the function is **not one-to-one**.

Visualized Solution

Understanding the Onto Condition

Forming the Quadratic in $x$

Condition for Real $x$

Simplifying the Discriminant

The $y$ -Quadratic Inequality

Solving for $α$

Finding the Interval for $α$

Testing One-to-One for $α = 3$

Conclusion on Injectivity

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

The Algebraic Bridge

The Discriminant Dance

The Global Inequality

The Final Verdict on Injectivity

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Visualized Solution

Understanding the Onto Condition

Forming the Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Condition for Real .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Simplifying the Discriminant

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-Quadratic Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding the Interval for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Testing One-to-One for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=3