Consider the function f:(0,2) defined by f(x)= x2+ 2x and the function g(x) defined by g(x)= cases \f(t)\, & 0<t x and 0<x 1\ 32+x, & 1<x<2 cases. Then,
- g is continuous but not differentiable at x=1
- g is continuous and differentiable for all x (0,2)
- g is not continuous for all x (0,2)
- g is neither continuous nor differentiable at x=1
Answer
(1) g is continuous but not differentiable at x=1
Solution
f'(x)= 12- 2x2<0 on (0,2) f is decreasing
0<t xf(t)=f(x), so g(x)= cases x2+ 2x, & 0<x 1\ 32+x, & 1<x<2 cases
[figure]
g(1-)=g(1)= 52=g(1+) g is continuous at x=1
LHD = 12-2=- 32, RHD =1 not differentiable at x=1 (corner)
Answer: (1)