The number of points, at which the function f(x)= \6x, 2+3x2 \+|x-1| |x2- 14 |, x (- , ), is not differentiable, is .
Answer
3
Solution
|x2- 14 |= (x2- 14 ) (cosine is even), which is differentiable everywhere.
(i) \6x, 2+3x2\: 6x=2+3x2 3x2-6x+2=0 x=1 1 3, both in (- , ).
At these points the slopes 6 and 6x 6 differ, so there are corners: 2 points.
For 1- 1 3<x<1+ 1 3 the maximum is 6x, so this part is differentiable at x=1.
(ii) |x-1| (x2- 14 ): at x=1 the factor 34 0, so there is a corner at x=1.
f is not differentiable at x=1- 1 3, 1, 1+ 1 3, i.e. at 3 points.
Answer: 3