# calculus

Find the values of c that satisfy the Mean Value Theorem for f(x)=6/x-3 on the interval [-1,2].

Is it no value of c in that interval because the function is not continuous on that interval???

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1. It is continuous between -1 and +3. The blow up is at x = 3.
y = 6/(x-3)
dy/dx = -6/(x-3)^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/(x-3)^2
(x-3)^2 = 4
x-3 = +/- 2
x = 5 or x = 1
only x = 1 is on the interval from -1 to 2

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