If $f(x) = |x|^3$,show that $f''(x)$ exists for all real $x$,and find it.

We know that the absolute value function is defined as:
$|x| = \begin{cases} x, & \text{if } x \ge 0 \\ -x, & \text{if } x < 0 \end{cases}$
Case $1$: When $x \ge 0$,$f(x) = |x|^3 = x^3$.
Then,$f'(x) = \frac{d}{dx}(x^3) = 3x^2$.
And $f''(x) = \frac{d}{dx}(3x^2) = 6x$.
Case $2$: When $x < 0$,$f(x) = |x|^3 = (-x)^3 = -x^3$.
Then,$f'(x) = \frac{d}{dx}(-x^3) = -3x^2$.
Wait,let us re-evaluate: $f(x) = |x|^3$. Since $|x|^3 = |x^3|$,if $x < 0$,$x^3 < 0$,so $|x^3| = -x^3$. Thus $f(x) = -x^3$ for $x < 0$.
$f'(x) = -3x^2$ for $x < 0$.
$f''(x) = -6x$ for $x < 0$.
However,let's check the continuity of $f'(x)$ at $x=0$:
$f'(0^+) = 3(0)^2 = 0$.
$f'(0^-) = -3(0)^2 = 0$.
Since $f'(0^+) = f'(0^-) = 0$,$f'(0) = 0$.
Now check $f''(x)$ at $x=0$:
$f''(0^+) = 6(0) = 0$.
$f''(0^-) = -6(0) = 0$.
Since $f''(0^+) = f''(0^-) = 0$,$f''(0) = 0$.
Thus,$f''(x) = 6|x|$ for all real $x$.