Question

If $f\left(x\right)=(x^2-4)|x^3-6x^2+11x-6|+\frac{x}{1+|x|}$, then the set of points at which the function $f\left(x\right)$ is not differentiable is?

• A. $\{-2,2,1,3\}$
• B. $\{-2,0,3\}$
• C. $\{-2,2,0\}$
• D. $\{1,3\}$

Answer 1

The correct option is D $\{1,3\}$

$f\left(x\right)=(x^2-4)|(x-1)(x^2-5x+6)|+\left(\frac{x}{1+|x|}\right)=(x^2-4)|(x-1)(x-2)(x-3)|+\left(\frac{x}{1+|x|}\right)$

|x| functions are not differentiable, at points where their value becomes zero because at these points it has sharp corners.

so at x=1, 2, 3 are possible points where it might not be differentiable but at x=2, it is multiplied by term $(x^2-4)$ which gives value zero and hence will neutralise the effect, so only at x={1, 3}, f(x) is not differentiable.