Let f 02 rightarrow R be differentiable with fO 0 fl 2 an

Let f : [0.2] rightarrow R be differentiable with f(O) = 0, f(l) = 2 and f(2) = 1. Prove that then exists a Provo that there exists a

Solution

According to the mean value theorem, if a function is continuous on the closed interval [a,b], in this case from [0,2], then there exists a differentiable function such that

f\'(c) = [f(b) - f(a)]/(b-a)

a) Since the function is continuous and differentiable in the closed range from [0,2], so there exists some c1 such that

f\'(c1) = [f(2) - f(0)]/(2-0) = (1-0)/(2-0) = 1/2

Hence the first part is proved that there exists a point c1, a < c1 < b, where f\'(c1) is equal to 1/2 or 0.5

b) Since the function is continuous and differentiable in the closed range from [1,2], so there exists some c3 such that

(*** It must be -1, since we can\'t apply MVT or mean value theorem to get the value of -1/2)

f\'(c3) = [f(2) - f(1)]/(2-1) = (1-2)/(2-1) = -1

Hence the first part is proved that there exists a point c3, a < c1 < b, where f\'(c3) is equal to -1

We can write that since the function is continuous and differentiable, hence the function differentiation will always be continuous

Hence there will be some value where it will be equal to -1/2

Since f\'(c1) = 1/2 and f\'(c3) = -1

So there will be one value where f\'(c2) = -1/2