Let g(x) be a linear function and f(x)= cases g(x) & , x 0 \ ( 1+x2+x ) 1x & , x>0 cases, is continuous at x=0. If f (1)=f(-1), then the value g(3) is
- e ( 49 )-1
- 13 e ( 49 e1/3 )
- e ( 49 e1/3 )
- 13 e ( 49 )+1
Answer
(3) e ( 49 e1/3 )
Solution
Let g(x)=ax+b. Continuity at 0: x 0+ ( 1+x2+x )1/x= ( 12 ) =0=g(0) ⇒ b=0
For x>0: e f= 1x e 1+x2+x
f'f=- 1x2 e 1+x2+x+ 1x ( 11+x- 12+x )
At x=1, f(1)= 23: f'(1)= 23 ( 16- e 23 )= 19- 23 e 23
f(-1)=g(-1)=-a a= 23 e 23- 19
g(3)=3a=2 e 23- 13
= e 49- e e1/3= e ( 49e1/3 )
Answer: (3)