rolle's theorem not applicable on f(x)=|x| in the interval [-1,1]

Question

a)f(x) not continuos in [-1,1]

b)f(x) not diferentiable in [-1,1]

c) f(-1) is not equal to f(1)

d) f(-1) is  equal to f(1)  but not equal to 0

Answer 1

Rolle's Theorem requires that the function it is being applied to be differentiable on the open interval (-1,1).  In this case, that's not true.  The absolute value function has a cusp at x=0, so it's not differentiable at that point.

 

The whole point of Rolle's Theorem is to say that the slope of the secant line through two points must be equal to the slope of the tangent line somewhere in between those two points, but that reasoning only works if the curve between your two points actually exists everywhere between your two points (continuous) and if the derivative also exists everywhere between your two points (differentiable).

 

So, let's compare it to g(x)=x^2.  In that case, we also have g(-1)=g(1)=1, but Rolle's theorem is true because y=x^2 is differentiable on its whole domain.  The slope of the secant line between x=-1 and x=1 is 0, so there must be a point on g where the slope of the tangent line is also 0 (and there is at x=0).  Going back to the absolute value function, f(-1)=f(1), and the slope of the secant line between those two points is 0.  But, there's no place between -1 and 1 where the slope of the tangent line is 0 because the derivative isn't continuous.  f'(x) just jumps from -1 to 1 at x=0, without going through a smooth transition like it does on y=x^2.  The reason is because f(x) has a cusp at x=0, where the derivative doesn't exist, while g(x) has a derivative at every point.

Answer 2

We can use the interval to evaluate f(x).

 

f(-1) = | -1 | = 1

f(0) = | 0 | = 0

f(1) = | 1 | = 1

 

Based on the calculations here, we can eliminate choice (a) and choice (c).  Any function defined by y is differentiable.

 

f(-1) = f(1) ≠ 0

 

This is true.

 

Your answer is (d).