Question

Let $f:R \rightarrow R$ be a function defined by $f(x) = |x^2-4|x|+3|$, then the number of points where $f(x)$ is non-differentiable is:

• A. 0
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

5

Answer 2

To find the number of points where $f(x) = |x^2-4|x|+3|$ is non-differentiable, we analyze the function by considering the definition of the absolute value and differentiability.

Let $g(x) = x^2-4|x|+3$. The function can be written as $f(x) = |g(x)|$. A function of the form $|h(x)|$ is non-differentiable at points where:

1. $h(x)=0$ and $h'(x) \ne 0$ at those points.
2. $h(x)$ itself is non-differentiable, and $h(x) \ne 0$ at those points.

First, let's analyze $g(x) = x^2-4|x|+3$. We can define $g(x)$ piecewise:

$g(x) = \begin{cases} x^2-4x+3 & \text{if } x \ge 0 \\ x^2+4x+3 & \text{if } x < 0 \end{cases}$

Case 1: Points where $g(x) = 0$.

If $x \ge 0$, $x^2-4x+3=0 \implies (x-1)(x-3)=0$. This gives $x=1$ and $x=3$. Both satisfy $x \ge 0$. If $x < 0$, $x^2+4x+3=0 \implies (x+1)(x+3)=0$. This gives $x=-1$ and $x=-3$. Both satisfy $x < 0$. So, $g(x)=0$ at $x \in \{-3, -1, 1, 3\}$.

Now, let's find $g'(x)$ for $x \ne 0$:

$g'(x) = \begin{cases} 2x-4 & \text{if } x > 0 \\ 2x+4 & \text{if } x < 0 \end{cases}$

Let's check $g'(x)$ at the points where $g(x)=0$:

• At $x=1$: $g'(1) = 2(1)-4 = -2$. Since $g(1)=0$ and $g'(1) \ne 0$, $f(x)$ is non-differentiable at $x=1$.
• At $x=3$: $g'(3) = 2(3)-4 = 2$. Since $g(3)=0$ and $g'(3) \ne 0$, $f(x)$ is non-differentiable at $x=3$.
• At $x=-1$: $g'(-1) = 2(-1)+4 = 2$. Since $g(-1)=0$ and $g'(-1) \ne 0$, $f(x)$ is non-differentiable at $x=-1$.
• At $x=-3$: $g'(-3) = 2(-3)+4 = -2$. Since $g(-3)=0$ and $g'(-3) \ne 0$, $f(x)$ is non-differentiable at $x=-3$.

So far, we have 4 points of non-differentiability: $x=-3, -1, 1, 3$.

Case 2: Points where $g(x)$ is non-differentiable.

The function $g(x)$ involves $|x|$, which is non-differentiable at $x=0$. So, we need to check differentiability of $g(x)$ at $x=0$.

Right-hand derivative of $g(x)$ at $x=0$: $g'(0^+) = \lim_{x \to 0^+} (2x-4) = -4$.

Left-hand derivative of $g(x)$ at $x=0$: $g'(0^-) = \lim_{x \to 0^-} (2x+4) = 4$.

Since $g'(0^+) \ne g'(0^-)$, $g(x)$ is non-differentiable at $x=0$.

Now, we check $f(x)=|g(x)|$ at $x=0$. At $x=0$, $g(0) = 0^2-4|0|+3 = 3$. Since $g(0)=3 \ne 0$, and $g(x)$ is continuous at $x=0$, $g(x)$ does not change sign around $x=0$. Specifically, $g(x)$ is positive in a neighborhood of $x=0$. Therefore, for $x$ in a neighborhood of $0$, $f(x) = |g(x)| = g(x)$. Since $f(x)=g(x)$ in a neighborhood of $x=0$, and $g(x)$ is non-differentiable at $x=0$, $f(x)$ is also non-differentiable at $x=0$.

Combining both cases, the points where $f(x)$ is non-differentiable are $x=-3, -1, 0, 1, 3$. There are 5 such points.