Question

Number of points where $f(x) = x^2 - |x^2 - 1| + 2||x| - 1| + 2|x| - 7$ is non differentiable is

Answer 1

0

Answer 2

The function $f(x) = x^2 - |x^2 - 1| + 2||x| - 1| + 2|x| - 7$ is an even function. Non-differentiability can occur at points where the arguments of absolute value functions are zero: $x^2 - 1 = 0 \implies x = \pm 1$; $|x| - 1 = 0 \implies x = \pm 1$; $|x| = 0 \implies x = 0$. The potential points of non-differentiability are $x = -1, 0, 1$.

Since $f(x)$ is even, we only need to check $x = 0$ and $x = 1$.

For $x \ge 0$, $|x| = x$. So, $f(x) = x^2 - |x^2 - 1| + 2|x - 1| + 2x - 7$.

Case 1: $0 \le x < 1$. $f(x) = x^2 - (1 - x^2) + 2(1 - x) + 2x - 7 = 2x^2 - 6$. The derivative is $f'(x) = 4x$. $f'(0^+) = 0$. $f'(1^-) = 4$.

Case 2: $x > 1$. $f(x) = x^2 - (x^2 - 1) + 2(x - 1) + 2x - 7 = 4x - 8$. The derivative is $f'(x) = 4$. $f'(1^+) = 4$.

At $x=0$: For $x < 0$, $|x| = -x$. $f(x) = x^2 - |x^2 - 1| + 2|-x - 1| + 2(-x) - 7$. For $-1 < x < 0$, $f(x) = x^2 - (1-x^2) + 2(x+1) - 2x - 7 = 2x^2 - 6$. The derivative is $f'(x) = 4x$. So, $f'(0^-) = 0$. Since $f'(0^-) = f'(0^+) = 0$, $f(x)$ is differentiable at $x=0$.

At $x=1$: Since $f'(1^-) = 4$ and $f'(1^+) = 4$, $f(x)$ is differentiable at $x=1$.

Since $f(x)$ is differentiable at $x=0$ and $x=1$, it is also differentiable at $x=-1$ due to symmetry. Therefore, the function is differentiable at all points $x = -1, 0, 1$. The number of non-differentiable points is 0.