Let f(x)= casesex-1, & x<0\ x2-5x+6, & x 0 cases and g(x)=f(|x|)+|f(x)|. If the number of points where g is not continuous and is not differentiable are and respectively, then + is equal to .
Answer
4
Solution
f(|x|)=|x|2-5|x|+6=x2-5|x|+6 for all x ( |x| 0)
For x<0: g(x)=x2+5x+6+ex-1
For x 0: g(x)= (x2-5x+6 )+ |x2-5x+6 |= cases2(x-2)(x-3), & 0 x 2\ 0, & 2<x<3\ 2(x-2)(x-3), & x 3 cases
Continuity at x=0: g(0-)=6+e-1, g(0)=12 discontinuous
Everywhere else g is continuous =1
Differentiability: not differentiable at x=0 (discontinuous).
At x=2: g'(2-)=2(2x-5) |x=2=-2, g'(2+)=0 not differentiable
At x=3: g'(3-)=0, g'(3+)=2(2x-5) |x=3=2 not differentiable
=3 (at x=0, 2, 3)
+ =1+3=4
Answer: 4