calculus
posted by Theresa .
Find the values of c that satisfy the Mean Value Theorem for f(x)=6/x3 on the interval [1,2].
Is it no value of c in that interval because the function is not continuous on that interval???

It is continuous between 1 and +3. The blow up is at x = 3.
y = 6/(x3)
dy/dx = 6/(x3)^2
now slope from 1 to +2
y(1) = 6/4 = 3/2
y(2) = 6/1 = 6
delta x = 2+1 = 3
so mean slope = (6 + 3/2)/3 = 2 + 1/2 = 1.5
now where does derivative = 1.5 ?
3/2 = 6/(x3)^2
(x3)^2 = 4
x3 = +/ 2
x = 5 or x = 1
only x = 1 is on the interval from 1 to 2
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