<p>Decide whether Rolle's Theorem can be applied to f(x) = [(x+2)(x-1)]/(x-1) on the interval [-2,0]. I know one of the conditions is that f(x) must be continuous on the interval [-2,0], which is true, and differentiable on the open interval (-2,0). f(x) would simplify to a line, f(x) = x+2 with a removable discontinuity at x=1. However, I'm not sure as to whether or not f(x) is differentiable at x=1. If there's a hole there, can it still be differentiated? Or would it not be differentiable and cause Rolle's Theorem to not be acceptable use here? Thanks.</p>
<p>If there is a hole at x=1, it is not differentiable.</p>
<p>Alright, thanks.</p>
<p>Another quick question, when it says "differentiable on the open interval (a,b)," that's just another way of saying all real numbers, right?</p>
<p>no (a,b) is (-2,0) therefore, rolles theorem works for this eq and interval since it is differentiable at all points where -2<x<0</p>
<p>"I know one of the conditions is that f(x) must be continuous on the interval [-2,0], which is true, and differentiable on the open interval (-2,0). f(x) would simplify to a line, f(x) = x+2 with a removable discontinuity at x=1. However, I'm not sure as to whether or not f(x) is differentiable at x=1."</p>
<p>I didn't really look at the numbers, but if there's a removable discontinuity at x=1, then your statement that the line was continuous was false. Because there's a discontinuity. Removable does not mean non-existent.</p>
<p>big mr pig...it only has to be continuous during the interval [a,b]...1 is not between -2 and 0</p>