Rolle’s TheoremMathematics 11: Lecture 22

Dan Sloughter

Furman University

October 22, 2007

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 1 / 9

Rolle’s Theorem

I If f is continuous on [a, b], differentiable on (a, b), and f (a) = f (b),then there exists a point c in (a, b) with f ′(c) = 0.

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 2 / 9

Proof

I Suppose f is continuous on the closed interval [a, b], differentiable onthe open interval (a, b), f (a) = f (b), and f ′(x) 6= 0 for all x in (a, b).

I By the extreme-value theorem, there exists a point c in [a, b] suchthat f (c) ≥ f (x) for all x in [a, b].

I By an earlier result, c cannot be in the open interval (a, b), and soeither c = a or c = b.

I Since f (a) = f (b), we have f (a) = f (b) ≥ f (x) for all x in [a, b].

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 3 / 9

Proof

I Suppose f is continuous on the closed interval [a, b], differentiable onthe open interval (a, b), f (a) = f (b), and f ′(x) 6= 0 for all x in (a, b).

I By the extreme-value theorem, there exists a point c in [a, b] suchthat f (c) ≥ f (x) for all x in [a, b].

I By an earlier result, c cannot be in the open interval (a, b), and soeither c = a or c = b.

I Since f (a) = f (b), we have f (a) = f (b) ≥ f (x) for all x in [a, b].

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 3 / 9

Proof

I Suppose f is continuous on the closed interval [a, b], differentiable onthe open interval (a, b), f (a) = f (b), and f ′(x) 6= 0 for all x in (a, b).

I By the extreme-value theorem, there exists a point c in [a, b] suchthat f (c) ≥ f (x) for all x in [a, b].

I By an earlier result, c cannot be in the open interval (a, b), and soeither c = a or c = b.

I Since f (a) = f (b), we have f (a) = f (b) ≥ f (x) for all x in [a, b].

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 3 / 9

Proof

I Suppose f is continuous on the closed interval [a, b], differentiable onthe open interval (a, b), f (a) = f (b), and f ′(x) 6= 0 for all x in (a, b).

I By the extreme-value theorem, there exists a point c in [a, b] suchthat f (c) ≥ f (x) for all x in [a, b].

I By an earlier result, c cannot be in the open interval (a, b), and soeither c = a or c = b.

I Since f (a) = f (b), we have f (a) = f (b) ≥ f (x) for all x in [a, b].

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 3 / 9

Proof (cont’d)

I Similarly, there exists a point d in [a, b] with f (d) ≤ f (x) for all x in[a, b].

I As before, it follows that d = a or d = b, and so f (a) = f (b) ≤ f (x)for all x in [a, b].

I Hence f (x) = f (a) = f (b) for all x in [a, b].

I That is, f is a constant function.

I But that implies f ′(x) = 0 for all x in (a, b), contradicting ourassumption that f ′(x) 6= 0 for all x in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 4 / 9

Proof (cont’d)

I Similarly, there exists a point d in [a, b] with f (d) ≤ f (x) for all x in[a, b].

I As before, it follows that d = a or d = b, and so f (a) = f (b) ≤ f (x)for all x in [a, b].

I Hence f (x) = f (a) = f (b) for all x in [a, b].

I That is, f is a constant function.

I But that implies f ′(x) = 0 for all x in (a, b), contradicting ourassumption that f ′(x) 6= 0 for all x in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 4 / 9

Proof (cont’d)

I Similarly, there exists a point d in [a, b] with f (d) ≤ f (x) for all x in[a, b].

I As before, it follows that d = a or d = b, and so f (a) = f (b) ≤ f (x)for all x in [a, b].

I Hence f (x) = f (a) = f (b) for all x in [a, b].

I That is, f is a constant function.

I But that implies f ′(x) = 0 for all x in (a, b), contradicting ourassumption that f ′(x) 6= 0 for all x in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 4 / 9

Proof (cont’d)

I Similarly, there exists a point d in [a, b] with f (d) ≤ f (x) for all x in[a, b].

I As before, it follows that d = a or d = b, and so f (a) = f (b) ≤ f (x)for all x in [a, b].

I Hence f (x) = f (a) = f (b) for all x in [a, b].

I That is, f is a constant function.

I But that implies f ′(x) = 0 for all x in (a, b), contradicting ourassumption that f ′(x) 6= 0 for all x in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 4 / 9

Proof (cont’d)

I Similarly, there exists a point d in [a, b] with f (d) ≤ f (x) for all x in[a, b].

I As before, it follows that d = a or d = b, and so f (a) = f (b) ≤ f (x)for all x in [a, b].

I Hence f (x) = f (a) = f (b) for all x in [a, b].

I That is, f is a constant function.

I But that implies f ′(x) = 0 for all x in (a, b), contradicting ourassumption that f ′(x) 6= 0 for all x in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 4 / 9

Application

I Suppose f is continuous on [a, b], differentiable on (a, b) withf ′(x) 6= 0 for all x in (a, b), and f (a) and f (b) are of opposite sign.

I By the Intermediate Value Theorem, there exists at least one point cin (a, b) for which f (c) = 0.

I By Rolle’s Theorem, there can exist at most one such point (because,if there were two, then there would be a point between them at whichthe derivative is 0).

I Hence the equation f (x) = 0 must have a unique solution in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 5 / 9

Application

I Suppose f is continuous on [a, b], differentiable on (a, b) withf ′(x) 6= 0 for all x in (a, b), and f (a) and f (b) are of opposite sign.

I By the Intermediate Value Theorem, there exists at least one point cin (a, b) for which f (c) = 0.

I By Rolle’s Theorem, there can exist at most one such point (because,if there were two, then there would be a point between them at whichthe derivative is 0).

I Hence the equation f (x) = 0 must have a unique solution in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 5 / 9

Application

I Suppose f is continuous on [a, b], differentiable on (a, b) withf ′(x) 6= 0 for all x in (a, b), and f (a) and f (b) are of opposite sign.

I By the Intermediate Value Theorem, there exists at least one point cin (a, b) for which f (c) = 0.

I By Rolle’s Theorem, there can exist at most one such point (because,if there were two, then there would be a point between them at whichthe derivative is 0).

I Hence the equation f (x) = 0 must have a unique solution in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 5 / 9

Application

I Suppose f is continuous on [a, b], differentiable on (a, b) withf ′(x) 6= 0 for all x in (a, b), and f (a) and f (b) are of opposite sign.

I By the Intermediate Value Theorem, there exists at least one point cin (a, b) for which f (c) = 0.

I By Rolle’s Theorem, there can exist at most one such point (because,if there were two, then there would be a point between them at whichthe derivative is 0).

I Hence the equation f (x) = 0 must have a unique solution in (a, b).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 5 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example

I Consider the equation 4x5 + x = 4.

I Note: If we let f (x) = 4x5 + x − 4, then the equation f (x) = 0 isequivalent to the equation 4x5 + x = 4.

I Now f (0) = −4 and f (1) = 1, so, by the Intermediate ValueTheorem, the equation 4x5 + x = 4 has at least one solution in theinterval (0, 1).

I Now f ′(x) = 20x4 + 1, and so f ′(x) ≥ 1 for all x .

I Hence, by Rolle’s Theorem, the equation 4x5 + x = 4 can have atmost one solution.

I It now follows that the equation has a unique solution, which lies inthe interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 6 / 9

Example (cont’d)

I Graph of f (x) = 4x5 + x − 4:

0.2 0.4 0.6 0.8 1

-4

-3

-2

-1

1

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 7 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example

I Consider the equation cos(x) = x .

I Note: If we let f (x) = x − cos(x), then the equation f (x) = 0 isequivalent to the equation cos(x) = x .

I Now f (0) = −1 and f (1) = 1− cos(1) > 0, so, by the IntermediateValue Theorem, the equation cos(x) = x has at least one solution inthe interval (0, 1).

I Moreover, f ′(x) = 1 + sin(x) > 0 for all x in (−1, 1).

I Hence f (x) = 0 has at most one solution in (−1, 1).

I Note: cos(x) 6= x for any x with |x | ≥ 1.

I It follows that the equation cos(x) = x has a unique solution, whichlies in the interval (0, 1).

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 8 / 9

Example (cont’d)

I Graphs of y = cos(x) and y = x :

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

Dan Sloughter (Furman University) Rolle’s Theorem October 22, 2007 9 / 9