if f(x) is twice differentiable defined for all x belongs to R such that f(x) =f(2-x) andf'(1/2)=f'(1/4) then
min no. of values in[0,2] for which f''(x) vanishes is

Question

if f(x) is twice differentiable defined for all x belongs to R such that f(x) =f(2-x) andf'(1/2)=f'(1/4) then
min no. of values in[0,2] for which f''(x) vanishes is

anonymous · Answer

emm...what that means?  "f''(x) vanishes"

anonymous · Answer

it means f''x=0

anonymous · Answer

f is symmetric with respect to x=1 and f''(x) vanishes once on the interval (0.25,0.5) so min number will b 2...but how to offer an acceptable reasoning

anonymous · Answer

$$f(x)=f(2-x)$$let  x=1-x$$f(1-x)=f(1+x)$$f is symmetric with respect to x=1

so u just need to work on [0,1]

anonymous · Answer

and i believe using mean value theorem for the interval [0.25,0.5] will work here

anonymous · Answer

can we find the period of the function

anonymous · Answer

i think it is 2

anonymous · Answer

but the function is defined for all real values

anonymous · Answer

i m confused whether the function is periodic or not

anonymous · Answer

for all real values

anonymous · Answer

oh sorry its not periodic

anonymous · Answer

just with considering f(x) =f(2-x) we cant tell its periodic$$f(x)=(x-1)^2$$