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

1. Domain check

Because of $\log|x|$, the function is only defined for $x \neq 0$.

2. Derivative gives the slope

3. Critical points (where slope is zero or undefined)

Set the derivative to zero:

So the real line (minus $0$) naturally splits into

4. Sign test on each interval

Take a simple test point in every interval and plug it into $4x - \tfrac1x$.

• Positive slope ⇒ increasing
• Negative slope ⇒ decreasing

Thus, the function is

• Increasing on $\big(-\tfrac12,0\big)$ and $(\tfrac12,\infty)$
• Decreasing on $(-\infty,-\tfrac12)$ and $(0,\tfrac12)$

What does the question ask?

We have a bumpy hill–valley road described by the rule

$y = 2x^2 \; - \; \log |x|$

The question wants to know where the road goes uphill (increasing) and where it goes downhill (decreasing).

How do we usually find that out?

1. Find the slope everywhere. The slope is just the derivative.
2. Check the sign of that slope.
   • Positive slope ⇒ going uphill (increasing).
   • Negative slope ⇒ going downhill (decreasing).
3. Split the road where it changes style (e.g.
   • at $x=0$ because $\log|x|$ breaks there,
   • and where the slope becomes zero).

That’s it! 😊

Step–by–Step Solution

1. Derivative

   $$\frac{dy}{dx}=4x - \frac1x\quad (x\neq0)$$

2. Solve $\dfrac{dy}{dx}=0$

   $$4x - \frac1x = 0 \;\Longrightarrow\; 4x^2 = 1 \;\Longrightarrow\; x = \pm\frac12$$

3. Test sign in each interval

   • For $x<-\tfrac12$: pick $x=-1$, $4(-1)-(-1)=-5<0$ (decreasing)
   • For $-\tfrac12<x<0$: pick $x=-\tfrac14$, $4(-\tfrac14)-(-4)=3>0$ (increasing)
   • For $0<x<\tfrac12$: pick $x=\tfrac14$, $1-4=-3<0$ (decreasing)
   • For $x>\tfrac12$: pick $x=1$, $4-1=3>0$ (increasing)

4. Write the final answer

   • Monotonically increasing for
     $\boxed{\,-\tfrac12 < x < 0\;\text{ and }\; x>\tfrac12\,}$
   • Monotonically decreasing for
     $\boxed{\,x<-\tfrac12\;\text{ and }\;0<x<\tfrac12\,}$