The number of critical points of the function f(x)= cases | xx |, & x 0\ 1, & x=0 cases in the interval (-2 , 2 ) is equal to :
- 1
- 3
- 5
- 7
Answer
(3) 5
Solution
Critical points are points where f'(x)=0 or f'(x) does not exist. f is even, so study x 0.
At x=0: near 0, f(x)= xx (positive, continuous since f(0)=1), differentiable with f'(0)=0 → critical point.
On (0, ): f(x)= xx, f'(x)= x x- xx2.
Let g(x)=x x- x: g(0)=0, g'(x)=-x x<0 ⇒ g<0 ⇒ f' 0 in (0, ).
At x= : f( )=0 is a minimum of | | with a sharp corner (left derivative - 1 , right derivative + 1 ) → f' does not exist → critical point.
On ( ,2 ): f(x)=- xx, f'(x)=0 x=x, which has exactly one root (x 4.49) in ( ,2 ) → critical point.
By symmetry the same happens at x=- and x -4.49.
[figure]
Critical points: 0, , 4.49 ⇒ total =5
Answer: (3)