Let f : R R be a differentiable function such that f ( x+y3 )= f(x)+f(y)3 for all x, y R, and f'(0)=3. Then the minimum value of the function g(x)=3+ex f(x), is:
- 3 ( e+1e )
- 3 ( e-1e )
- 3-ee
- 3e
Answer
(2) 3 ( e-1e )
Solution
x=y=0: f(0)= 2f(0)3 f(0)=0
y=0: f ( x3 )= f(x)3
f(x+y)3=f ( x+y3 )= f(x)+f(y)3 f(x+y)=f(x)+f(y)
f'(x)= h 0 f(x+h)-f(x)h= h 0 f(h)-f(0)h=f'(0)=3
f(x)=3x ( f(0)=0)
g(x)=3+3xex, g'(x)=3ex(1+x)
g'(x)<0 for x<-1 and g'(x)>0 for x>-1 minimum at x=-1
g =g(-1)=3- 3e=3 ( e-1e )
Answer: (2)