Determine the intervals of monotonicity of the function y

Determine the intervals of monotonicity of the function $y = x^{2}-\ln x^{2}$.

Step 1: Domain and symmetry

Since $\ln x^{2} = 2\ln|x|$, the function may be written $y = x^{2}-2\ln|x|$, defined for $x\neq0$. It is even: \[y(-x) = \bigl(-x\bigr)^{2}-\ln\bigl(-x\bigr)^{2} = x^{2}-\ln x^{2} = y(x),\]

so the left half of the graph is the mirror image of the right half.

Step 2: The derivative

\[ y' = 2x-\frac2x = \frac{2x^{2}-2}{x} = \frac{2\bigl(x-1\bigr)\bigl(x+1\bigr)}{x}. \tag{1} \]

Step 3: Critical points and sign

$y' = 0$ at $x = \pm1$, and $y'$ is undefined at $x = 0$, which is outside the domain. The three sign changes split the domain into four intervals: \[x\lt -1:\ \frac{(-)(-)}{(-)} = -\ (\searrow); \qquad -1\lt x\lt 0:\ \frac{(-)(+)}{(-)} = +\ (\nearrow);\]

\[ 0\lt x\lt 1:\ \frac{(-)(+)}{(+)} = -\ (\searrow); \qquad x\gt 1:\ \frac{(+)(+)}{(+)} = +\ (\nearrow). \]

Step 4: Classify the critical points

At both $x = -1$ and $x = 1$ the derivative changes from $-$ to $+$: each is a local minimum, and by evenness they carry the same value

\[ y(\pm1) = 1-\ln1 = 1. \]

Step 5: Verification

\[y(0.5) = 1.6362944 \gt y(1) = 1 \lt y(2) = 2.6137056 \;\;\checkmark\]

\[y(-0.5) = 1.6362944 = y(0.5),\ \ y(-1) = 1 = y(1) \;\;\checkmark\ \text{(evenness)}\]

\[y'' = 2+\frac{2}{x^{2}} \gt 0 \text{ everywhere} \;\Longrightarrow\; \text{both critical points are minima} \;\;\checkmark\]