Which of the following functions $f(x)$ are differentiable at $x = 0$?

(A) $|x|$

(B) $|x - 1|$

(C) $[x]$, where $[t]$ denotes the greatest integer $\leq t$

(D) $|x + 1|$

(E) $x^2$

Choose the correct answer from the options given below:

A function is differentiable at $x = 0$ if the left-hand derivative and right-hand derivative both exist and are equal at that point.

---

For (A) $f(x) = |x|$:

At $x = 0$: $f(0) = |0| = 0$

The right-hand derivative (approaching from positive side): slope $= +1$

The left-hand derivative (approaching from negative side): slope $= -1$

Since the left and right derivatives are not equal, $|x|$ is not differentiable at $x = 0$.

---

For (B) $f(x) = |x - 1|$:

At $x = 0$: $f(0) = |0 - 1| = 1$

The corner of $|x - 1|$ occurs at $x = 1$, not at $x = 0$.

For $x$ near $0$: Since $x < 1$, we have $x - 1 < 0$, so $|x - 1| = -(x - 1) = 1 - x$

The derivative is $f'(x) = -1$, which exists at $x = 0$.

Therefore $|x - 1|$ is differentiable at $x = 0$.

---

For (C) $f(x) = [x]$ (greatest integer function):

At $x = 0$: $f(0) = [0] = 0$

The greatest integer function has a jump discontinuity at $x = 0$ (jumps from $-1$ to $0$).

Since the function is not continuous at $x = 0$, it cannot be differentiable there.

---

For (D) $f(x) = |x + 1|$:

At $x = 0$: $f(0) = |0 + 1| = 1$

The corner of $|x + 1|$ occurs at $x = -1$, not at $x = 0$.

For $x$ near $0$: Since $x > -1$, we have $x + 1 > 0$, so $|x + 1| = x + 1$

The derivative is $f'(x) = 1$, which exists at $x = 0$.

Therefore $|x + 1|$ is differentiable at $x = 0$.

---

For (E) $f(x) = x^2$:

Polynomial functions are differentiable everywhere.

The derivative is $f'(x) = 2x$, so $f'(0) = 0$ exists.

Therefore $x^2$ is differentiable at $x = 0$.

---

The functions differentiable at $x = 0$ are (B), (D), and (E).