How do you verify whether rolle's theorem can be applied to the function #f(x)=1/x^2# in [-1,1]?

Answer 1

#f(x)# does not satisfy the conditions of Rolle's Theorem on the interval #[-1,1]#, as #f(x)# is not continuous on the interval

Rolle's Theorem states that if a function, #f(x)# is continuous on the closed interval #[a,b]#, and is differentiable on the interval, and #f(a)=f(b)#, then there exists at least one number #c#, in the interval such that #f'(c)=0#.

So what Rolle's Theorem is stating should be obvious as if the function is differentiable then it must be continuous (as differentiability #=># continuity), and if its is continuous and #f(a)=f(b)# then the curve must chge direction (at least once) so it must have at least one minimum or maximum in the interval.

With #f(x)=1/x^2# with #x in [-1,1]#, then we can see #f(1)=1=f(-1)#, so our quest to verify Rolle's Theorem is to find a number #c# st #f'(c)=0#.

Differentiating wrt #x# we have, #f'(x)=-2x^-3=-(2)/x^3#

To find a turning point we require;

# f'(x)=0 => -(2)/x^3 = 0#

Which has no finite solution. We can conclude that #f(x)# does not satisfy the conditions of Rolle's Theorem on the interval #[-1,1]#, so it must be that #f(x)# is not continuous in that interval.

We can see that this is the case graphically, as f(x) has a discontinuity when #x=0 in [-1,1]#:

graph{1/x^2 [-10, 10, -2, 10]}

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Answer 2

To verify whether Rolle's Theorem can be applied to the function (f(x) = \frac{1}{x^2}) in the interval ([-1, 1]), follow these steps:

  1. Check if the function is continuous on the closed interval ([-1, 1]). Since (f(x) = \frac{1}{x^2}) is continuous everywhere on its domain except at (x = 0), and ([-1, 1]) does not include (x = 0), the function is continuous on ([-1, 1]).

  2. Check if the function is differentiable on the open interval ((-1, 1)). Since (f(x) = \frac{1}{x^2}) is differentiable everywhere on its domain except at (x = 0), and ([-1, 1]) does not include (x = 0), the function is differentiable on ((-1, 1)).

  3. Verify if (f(-1) = f(1)). Evaluate (f(-1)) and (f(1)). If (f(-1) = f(1)), then Rolle's Theorem can be applied.

Let's check: [f(-1) = \frac{1}{(-1)^2} = 1] [f(1) = \frac{1}{(1)^2} = 1]

Since (f(-1) = f(1)), Rolle's Theorem can be applied to the function (f(x) = \frac{1}{x^2}) in the interval ([-1, 1]).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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