Find a number c which satisfies the conclusions of the mean value theorem for the given functions & intervals?

i) f(x) = x^3; [1, 2]


ii) f(x) = sqrt(x); [0, 2]





B) Let f(x) = 1/x: Show that there is no c∈[-1, 2] such that


f'(c) =[ f(2) - f(-1) ] / [2 - (-1)]





Why does this not contradict the mean value theorem?

Find a number c which satisfies the conclusions of the mean value theorem for the given functions %26amp; intervals?
i) f(x)=x^3 [1,2]. f is a polynomial so it is continuous and differentiable. Hence, the MVT states that:





f(2)-f(1) = f'(c)(2-1) for some c between 1 and 2.





f(2)=8


f(1)=1. Hence, you have: 8-1=3c^2(1), or 7=3c^2. hence c^2=7/3 and c=sqrt(7/3) or c=-sqrt(7/3). Since c is between 1 and 2, c=sqrt(7/3)





ii) f(2)=sqrt(2)


f(0)=0





Hence, you have: sqrt(2)-0=1/2sqrt(c) (2) or sqrt(2)=1/sqrt(c).





This means that sqrt(2c)=1 or 2c=1, thus c=1/2.





B) f(x) is not continuous nor differentiable on [-1,2] (as it is undefined at x=0)