The function f(x) = x / (x² - 6x - 16), x ∈ R - {-2, 8} (1) decreases in (-2, 8) and increases in (-

Q: Wait, why do we have to exclude $-2$ and $8$ from the intervals? Can't we just say it decreases everywhere?

That is a great observation! While the derivative is negative everywhere it is defined, the function itself is not continuous at $x = -2$ and $x = 8$. Because the function is undefined at these points, we cannot say it decreases on the entire real line. We must express the interval of decrease as the union of the three distinct regions: $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$.

Q: Why can't we just use the sign scheme on the numerator and denominator to find the intervals?

You absolutely can! However, in this specific case, the denominator $(x^2 - 6x - 16)^2$ is a perfect square, meaning it is always positive for all $x$ in the domain. Similarly, the numerator $-(x^2 + 16)$ is always negative. Since a negative divided by a positive is always negative, the derivative is negative across its entire domain, which simplifies the sign scheme significantly.

Q: Is it possible for a function to be decreasing everywhere in its domain?

Yes, absolutely! A function is strictly decreasing on an interval if its derivative is strictly negative on that interval. In this problem, because $f'(x) < 0$ for all $x$ in the domain, the function is strictly decreasing on each of the three intervals defined by the asymptotes. It is a classic example of a function that never 'turns' to increase.

Question

The function $f(x) = \frac{x}{x^2 - 6x - 16}, x \in \mathbb{R} - \{-2, 8\}$

Options:

(A) decreases in $(-2, 8)$ and increases in $(-\infty, -2) \cup (8, \infty)$

(B) decreases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$

(C) decreases in $(-\infty, -2)$ and increases in $(8, \infty)$

(D) increases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$

Sigma Percentile · Accepted Answer

### The Rollercoaster of Rational Functions

Imagine you are standing before the graph of $f(x) = \frac{x}{x^2 - 6x - 16}$. At first glance, it looks like a complex, intimidating rational function.

But in the world of JEE Advanced, we don't fear functions; we dissect them. Let's embark on a journey to understand the monotonicity of this curve.

### Phase 1: Mapping the Forbidden Zones

Before we even touch calculus, we must respect the domain. A rational function is only as strong as its denominator.

We factor the denominator: 
$$x^2 - 6x - 16 = (x - 8)(x + 2)$$

This reveals the 'forbidden zones'—the values $x = -2$ and $x = 8$. At these points, the denominator vanishes, and the function explodes toward infinity, creating vertical asymptotes.

Our domain is $\mathbb{R} - \{-2, 8\}$. Understanding these boundaries is crucial because they define the 'rooms' in which our function lives.

### Phase 2: The Calculus of Slopes

To find where the function increases or decreases, we need to know the slope of the tangent line at any point. This is the job of the first derivative, $f'(x)$.

Since our function is a ratio, we deploy the Quotient Rule: 
$$\frac{d}{dx}\left(\frac{u}{v}ight) = \frac{v u' - u v'}{v^2}$$

Let $u = x$ (so $u' = 1$) and $v = x^2 - 6x - 16$ (so $v' = 2x - 6$). Now, we carefully assemble the pieces:

$$f'(x) = \frac{(x^2 - 6x - 16)(1) - (x)(2x - 6)}{(x^2 - 6x - 16)^2}$$

### Phase 3: The Algebraic Cleanup

This is where many students stumble, but let's stay focused. We expand the numerator:

$$	ext{Numerator} = x^2 - 6x - 16 - (2x^2 - 6x)$$

Distributing that negative sign is the most common place for a 'silly mistake.' Let's do it with precision:

$$	ext{Numerator} = x^2 - 6x - 16 - 2x^2 + 6x$$

Notice the beauty of the cancellation: $-6x$ and $+6x$ vanish into thin air. We are left with $x^2 - 2x^2 - 16$, which simplifies to $-x^2 - 16$.

Factoring out the negative sign, we get $-(x^2 + 16)$. Our derivative is now:

$$f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}$$

### Phase 4: The Final Verdict

Now, look at the expression. The term $(x^2 + 16)$ is always positive for any real $x$. The denominator is a squared term, so it is also always positive.

With a negative sign sitting in front of the numerator, the entire derivative $f'(x)$ is **strictly negative** for all $x$ in the domain.

A negative derivative means the slope is always pointing downward. The function is **strictly decreasing**.

Because the function is broken into three parts by the asymptotes at $x = -2$ and $x = 8$, we conclude that the function decreases in the union of these intervals: 
$$(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$$

You have just successfully analyzed the behavior of a rational function. Keep this analytical mindset, and no function will ever be too complex for you to master!

Animated Solution for Mathematics - Differentiation: The function $f (x) = \frac{x}{x ^{2} - 6 x - 16}, x \in R - {- 2, 8}$

Select Answer:

Visualized Solution

Function and Domain of $f (x)$

The Monotonicity Test for $f (x)$

Quotient Rule Setup for $f^{'} (x)$

Applying the Quotient Rule

Expanding the Numerator

Simplifying the Numerator

Final Form of $f^{'} (x)$

Sign Analysis of $f^{'} (x)$

Conclusion on Monotonicity

Visualizing the Decreasing $f (x)$

The Sigma Insight: Monotonicity

The Rollercoaster of Rational Functions

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=x2−6x−16x​,x∈R−{−2,8}

Select Answer:

Visualized Solution

Function and Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

The Monotonicity Test for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Quotient Rule Setup for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Applying the Quotient Rule

Expanding the Numerator

Simplifying the Numerator

Final Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Sign Analysis of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Conclusion on Monotonicity

Visualizing the Decreasing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)