analysis 2 5 Let f 02 2 an a Prove that there exists a ci E

5. Let f (0,2 2 an (a) Prove that there exists a ci E 0,2 with J (ci) t. (b) Prove that there exists a E]0,2 with f\'(c2)

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 c2 such that

f\'(c2) = [f(2) - f(1)]/(2-1) = (1-2)/(2-1) = -1

Hence the first part is proved that there exists a point c2, a < c1 < b, where f\'(c2) is equal to -1