Mean value theorem 2

g(x)=f(x)-2x

dude, let the rookies try them. I dont want an answer but the approach is more important

f is differentiable between[0,2].Hence also differentiable in interval[0,1],[1,2]
for [0,1]
f(0)=0,f(1)=2
therefore there must be 1 value of x in(0,1) where f'(x)=f(1)-f(0)/1-0=2
therefore in (0,1) at a pt. f'(x)=2
similarly for [1,2]
we get another value of x in (1,2) where f'(x)=f(2)-f(1)/2-1=2
therefore in [0,2] there are 2 values of x for which f'(x)=2.Let them be x1,x2
now plot a graph of f'(x) vs x.
f'(x) is continuous and f''(x) exists .
there must be 1 value of x in between x1,x2 where f''(x)=f'(x2)-f'(x1)/x2-x1=0
x1,x2 lies in (0,2)
therfore that x also lies in (0,2).[proved]

g(x)=f(x)-2x

g(0)=g(1)=g(2)

So for some λ in (0,1), g'(x)=0

Again for some α in (1,2), g'(x)=0

From which for some ξ in (λ,α), g"(x)=f"(x)=0