The function y = 2x² - ln |x| is monotonically increasing for values of x(≠ 0) satisfying the inequa

Q: Wait, why is $x=0$ a critical point if it is not in the domain?

That is a fantastic observation! A critical point is defined as any point where the derivative is either zero or undefined. Even though $x=0$ is not in the domain of the original function, it is a point where the derivative $\frac{dy}{dx}$ is undefined. Because the function's behavior changes across this asymptote, we must include it in our sign scheme to correctly identify the intervals of monotonicity.

Q: Can I just ignore the absolute value sign in $\ln |x|$?

You cannot simply ignore it, but you can interpret it! The absolute value $|x|$ means that for $x > 0$, $\ln |x| = \ln(x)$, and for $x < 0$, $\ln |x| = \ln(-x)$. Interestingly, the derivative of $\ln |x|$ is $\frac{1}{x}$ in both cases. So, while the absolute value is crucial for the domain, the derivative simplifies beautifully to $\frac{1}{x}$ regardless of the sign of $x$.

Q: Why does the sign alternate at $x=0$ even though it is not a root of the numerator?

Great question! The sign of the derivative $\frac{dy}{dx} = \frac{(2x-1)(2x+1)}{x}$ depends on all three factors: $(2x-1)$, $(2x+1)$, and $x$. When we cross $x=0$, we are crossing a factor in the denominator with an odd exponent (power of 1). In the Wavy Curve Method, crossing a root or an asymptote with an odd exponent always causes the sign of the entire expression to flip. That is why the sign alternates at $x=0$!

Question

The function $y = 2x^2 - \ln |x|$ is monotonically increasing for values of $x(
eq 0)$ satisfying the inequalities ...... and monotonically decreasing for values of $x$ satisfying the inequalities .........

Sigma Percentile · Accepted Answer

### Analyzing the Domain

To understand the behavior of the function $y = 2x^2 - \ln |x|$, we must first respect the gatekeeper of the function: the domain. Because the function involves a natural logarithm, the argument $|x|$ cannot be zero.

Thus, our domain is defined as $x \in \mathbb{R} \setminus \{0\}$. This is not merely a technicality; it is a physical boundary that dictates the structure of our entire analysis.

### The Engine of Change

To determine where the function is increasing or decreasing, we must calculate the derivative, $\frac{dy}{dx}$. Applying the power rule to $2x^2$ yields $4x$, and the derivative of $\ln |x|$ is the classic result $\frac{1}{x}$.

Combining these, we obtain the derivative:

$$ \frac{dy}{dx} = 4x - \frac{1}{x} $$

To facilitate further analysis, we combine these terms into a single rational expression:

$$ \frac{dy}{dx} = \frac{4x^2 - 1}{x} $$

This expression serves as the engine of our analysis, allowing us to observe exactly how the slope behaves across the domain.

### The Wavy Curve Method

We identify the critical points where the derivative is zero or undefined. Setting the numerator $4x^2 - 1 = 0$ yields $x = \pm \frac{1}{2}$, while setting the denominator to zero yields $x = 0$.

These three points—$-\frac{1}{2}$, $0$, and $\frac{1}{2}$—act as the markers on our map. We now apply the Wavy Curve Method to determine the sign of the derivative in each interval by analyzing the expression:

$$ \frac{dy}{dx} = \frac{(2x - 1)(2x + 1)}{x} $$

For $x > \frac{1}{2}$, all factors are positive, so the derivative is positive and the function is **increasing**. As we cross the root at $x = \frac{1}{2}$, the sign flips to negative, indicating the function is **decreasing** on the interval $(0, \frac{1}{2})$.

Crossing the asymptote at $x = 0$ flips the sign again to positive on $(-\frac{1}{2}, 0)$. Finally, crossing $x = -\frac{1}{2}$ makes the derivative negative on the interval $(-\infty, -\frac{1}{2})$.

### The Final Revelation

By carefully tracking these sign changes, we arrive at our conclusion regarding the monotonicity of the function.

The function is **monotonically increasing** where $\frac{dy}{dx} > 0$, which occurs in the intervals:

$$ x \in \left( -\frac{1}{2}, 0 \right) \cup \left( \frac{1}{2}, \infty \right) $$

Conversely, the function is **monotonically decreasing** where $\frac{dy}{dx} < 0$, which occurs in the intervals:

$$ x \in \left( -\infty, -\frac{1}{2} \right) \cup \left( 0, \frac{1}{2} \right) $$

You have successfully navigated the slopes of this function using the power of calculus. Remember that every such problem allows you to visualize the hidden geometry of the mathematical world.

Animated Solution for Mathematics - Differentiation: The function $y = 2 x^{2} - ln ∣ x ∣$ is monotonically increasing for values of $x (\neq = 0)$ satisfying the inequalities ...... and monotonically decreasing for values of $x$ satisfying the inequalities .........

Visualized Solution

Function $y = 2 x^{2} - ln ∣ x ∣$

Monotonicity Condition

Differentiating the Function

Simplifying $\frac{d y}{d x}$

Factorizing the Numerator

Finding Critical Points

Sign Scheme Analysis

Interval Signs: $x > 0$

Interval Signs: $x < 0$

Final Intervals

The Sigma Insight: Monotonicity

Analyzing the Domain

The Engine of Change

The Wavy Curve Method

The Final Revelation

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Visualized Solution

Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2x2−ln∣x∣

Monotonicity Condition

Differentiating the Function

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Factorizing the Numerator

Finding Critical Points

Sign Scheme Analysis

Interval Signs: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0

Interval Signs: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Final Intervals

The Sigma Insight: Monotonicity

Analyzing the Domain

The Engine of Change

The Wavy Curve Method

The Final Revelation

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