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

The function is $f(x) = (x^2 - x - 2) \, |x^2 + x - 6|$. The absolute value makes the function piecewise, and the derivative may fail to exist at points where the expression inside the absolute value is zero — because that's where the "kink" or corner can appear. But we also need to check whether the factor outside the absolute value cancels that kink, making the function smooth there.

Let’s find the critical points.

1. Factor everything.

   $x^2 - x - 2 = (x - 2)(x + 1)$

   $x^2 + x - 6 = (x + 3)(x - 2)$

   So $f(x) = (x - 2)(x + 1) \, |(x + 3)(x - 2)|$.

2. Identify where the absolute value changes behaviour.

   The expression inside the absolute value is zero at $x = -3$ and $x = 2$. These are the only points where the definition of $| \cdot |$ switches between $+$ and $-$.

3. Check differentiability at $x = -3$.

   Near $x = -3$, the factor $(x - 2)$ is non-zero, and $(x + 1)$ is non-zero. The absolute value part behaves like $|(x+3)(x-2)|$. Since $(x-2)$ is non-zero and constant in sign near $-3$, the kink from $|x+3|$ survives. The product $(x-2)(x+1)$ is non-zero at $x=-3$, so the function has a corner there — the left and right derivatives will differ. Hence $f$ is not differentiable at $x = -3$.

4. Check differentiability at $x = 2$.

   Here the factor $(x-2)$ appears both outside the absolute value and inside it. Write $f(x) = (x-2)(x+1) \, |(x+3)(x-2)|$.

   For $x$ near 2, $(x+3)$ is positive, so $|(x+3)(x-2)| = (x+3)|x-2|$.

   Thus $f(x) = (x-2)(x+1)(x+3)|x-2| = (x+1)(x+3) \, (x-2)|x-2|$. …

The key idea here is that an absolute value function $|x - a|$ has a sharp corner (a cusp) at $x = a$, where its left-hand and right-hand derivatives differ. When you add two such functions, the sum inherits the non-differentiability at each of those corner points, unless the corners somehow cancel each other out — which they don't here.

Let’s see why step by step.

1. Identify the critical points.

   The function $f(x) = |x+1| + |x-1|$ has two absolute value expressions. The first, $|x+1|$, changes its behaviour at $x = -1$. The second, $|x-1|$, changes at $x = 1$. So the natural points to check for differentiability are $x = -1$ and $x = 1$.

2. Write the piecewise definition.

   For $x < -1$, both $x+1$ and $x-1$ are negative, so $|x+1| = -(x+1)$ and $|x-1| = -(x-1)$.

   For $-1 \le x < 1$, $x+1 \ge 0$ but $x-1 < 0$, so $|x+1| = x+1$ and $|x-1| = -(x-1)$.

   For $x \ge 1$, both are non-negative, so $|x+1| = x+1$ and $|x-1| = x-1$.

   This gives:

3. Check differentiability at $x = -1$.

   The left-hand derivative (from $x < -1$) is the derivative of $-2x$, which is $-2$.

   The right-hand derivative (from $x > -1$, but still $x < 1$) is the derivative of the constant $2$, which is $0$.

   Since $-2 \neq 0$, the left and right derivatives are different. Therefore $f$ is not differentiable at $x = -1$.

4. Check differentiability at $x = 1$.

   The left-hand derivative (from $x < 1$, but $x > -1$) is the derivative of the constant $2$, which is $0$.

   The right-hand derivative (from $x > 1$) is the derivative of $2x$, which is $2$.

   Again, $0 \neq 2$, so $f$ is not differentiable at $x = 1$.

5. Check any other point.

   For $x < -1$, $f(x) = -2x$ is a straight line, differentiable everywhere in that open interval. …

Concept. A piecewise function can only lose continuity or differentiability where its defining formula changes or blows up — here, only at $x = 0$. Everywhere else $f(x) = x^2\cos\dfrac{\pi}{x}$ is a product of the polynomial $x^2$ with $\cos(\pi/x)$, both infinitely differentiable for $x \ne 0$.

Step 1 — the point in question. $x = 2$ lies in the open interval $(1,3)$, which excludes $0$; on this interval $f$ is the single smooth expression $x^2\cos(\pi/x)$.

Step 2 — compute the derivative. For $x \ne 0$: …

Substitution. Put $x=3-h$ with $h\to0^+$. Then

The expression reduces to $\dfrac{(1+\sin h)\cdot 1}{h\cdot(-1)}=\dfrac{1+\sin h}{-h}$. …