Rolle's theorem, mean value theorem
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1. Rolle's theorem states that if a function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), and f(a) = f(b), then there exists at least one number c in (a,b) where the derivative f'(c) = 0. 2. The mean value theorem states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), there exists a number c in (a,b) such that f'(c) = (f(b) - f(a))/(b - a).
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Lecture 6 : Rolle’s Theorem, Mean Value Theorem
The reader must be familiar with the classical maxima and minima problems from calculus. For
example, the graph of a differentiable function has a horizontal tangent at a maximum or minimum
point. This is not quite accurate as we will see.
Definition : Let f : I → R, I an interval. A point x0 ∈ I is a local maximum of f if there is
a δ > 0 such that f (x) ≤ f (x0 ) whenever x ∈ I ∩ (x0 − δ, x0 + δ). Similarly, we can define local
minimum.
Theorem 6.1 : Suppose f : [a, b] → R and suppose f has either a local maximum or a local
minimum at x0 ∈ (a, b). If f is differentiable at x0 then f (x0 ) = 0.
Proof: Suppose f has a local maximum at x0 ∈ (a, b). For small (enough) h, f (x0 + h) ≤ f (x0 ).
If h > 0 then
f (x0 + h) − f (x0 )
≤ 0.
h
Similarly, if h < 0, then
f (x0 + h) − f (x0 )
≥ 0.
h
By elementary properties of the limit, it follows that f (x0 ) = 0.
We remark that the previous theorem is not valid if x0 is a or b. For example, if we consider the
function f : [0, 1] → R such that f (x) = x, then f has maximum at 1 but f (x) = 1 for all x ∈ [0, 1].
The following theorem is known as Rolle’s theorem which is an application of the previous
theorem.
Theorem 6.2 : Let f be continuous on [a, b], a < b, and differentiable on (a, b). Suppose f (a) =
f (b). Then there exists c such that c ∈ (a, b) and f (c) = 0.
Proof: If f is constant on [a, b] then f (c) = 0 for all c ∈ [a, b]. Suppose there exists x ∈ (a, b) such
that f (x) > f (a). (A similar argument can be given if f (x) < f (a)). Then there exists c ∈ (a, b)
such that f (c) is a maximum. Hence by the previous theorem, we have f (c) = 0.
Problem 1 : Show that the equation x13 + 7x3 − 5 = 0 has exactly one (real) root.
Solution : Let f (x) = x13 + 7x3 − 5. Then f (0) < 0 and f (1) > 0. By the IVP there is at least one
positive root of f (x) = 0. If there are two distinct positive roots, then by Rolle’s theorem there is
some x0 > 0 such that f (x0 ) = 0 which is not true. Moreover, observe that f (x) < 0 for x < 0.
Problem 2 : Let f and g be functions, continuous on [a, b], differentiable on (a, b) and let f (a) =
f (b) = 0. Prove that there is a point c ∈ (a, b) such that g (c)f (c) + f (c) = 0.
Solution : Define h(x) = f (x)eg(x) . Here, h(x) is continuous on [a, b] and differentiable on (a, b).
Since h(a) = h(b) = 0, by Rolle’s theorem, there exists c ∈ (a, b) such that h (c) = 0.
Since h (x) = [f (x)+g (x)f (x)]eg(x) and eα = 0 for any α ∈ R, we see that f (c)+g (c)f (c) = 0.
A geometric interpretation of the above theorem can be given as follows. If the values of a
differentiable function f at the end points a and b are equal then somewhere between a and b there
is a horizontal tangent. It is natural to ask the following question. If the value of f at the end
points a and b are not the same, is it true that there is some c ∈ [a, b] such that the tangent line
at c is parallel to the line connecting the endpoints of the curve? The answer is yes and this is
essentially the Mean Value Theorem.](https://image.slidesharecdn.com/rollestheoremmeanvaluetheorem-121127120402-phpapp02/85/Rolle-s-theorem-mean-value-theorem-1-320.jpg)
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Theorem 6.3 : (Mean Value Theorem) Let f be continuous on [a, b] and differentiable on
(a, b). Then there exists c ∈ (a, b) such that f (b) − f (a) = f (c)(b − a).
Proof: Let
f (b) − f (a)
g(x) = f (x) − (x − a).
b−a
Then g(a) = g(b) = f (a). The result follows by applying Rolle’s Theorem to g.
The mean value theorem is an important result in calculus and has some important applications
relating the behaviour of f and f . For example, if we have a property of f and we want to see
the effect of this property on f , we usually try to apply the mean value theorem. Let us see some
examples.
Example 1 : Let f : [a, b] → R be differentiable. Then f is constant if and only if f (x) = 0 for
every x ∈ [a, b].
Proof : Suppose that f is constant, then from the definition of f (x) it is immediate that f (x) = 0
for every x ∈ [a, b].
To prove the converse, let a < x ≤ b. By the mean value theorem there exists c ∈ (a, x) such
that f (x)−f (a) = f (c)(x−a). Since f (c) = 0, we conclude that f (x) = f (a), that is f is constant.
(If we try to prove the converse directly from the definition of f (x) we will be in trouble.)
Example 2 : Suppose f is continuous on [a, b] and differentiable on (a, b).
(i) If f (x) = 0 for all x ∈ (a, b), then f is one-one (i.e, f (x) = f (y) whenever x = y).
(ii) If f (x) ≥ 0 (resp. f (x) > 0) for all x ∈ (a, b) then f is increasing (resp. strictly increasing)
on [a, b]. (We have a similar result for decreasing functions.)
Proof : Apply the mean value theorem as we did in the previous example. (Note that f can be
one-one but f can be 0 at some point, for example take f (x) = x3 and x = 0.)
Problem 3 : Use the mean value theorem to prove that | sinx − siny | ≤ | x − y | for all x, y ∈ R.
Solution : Let x, y ∈ R. By the mean value theorem sinx − siny = cosc (x − y) for some c between
x and y. Hence | sinx − siny | ≤ | x − y |.
Problem 4 : Let f be twice differentiable on [0, 2]. Show that if f (0) = 0, f (1) = 2 and f (2) = 4,
then there is x0 ∈ (0, 2) such that f (x0 ) = 0.
Solution : By the mean value theorem there exist x1 ∈ (0, 1) and x2 ∈ (1, 2) such that
f (x1 ) = f (1) − f (0) = 2 and f (x2 ) = f (2) − f (1) = 2.
Apply Rolle’s theorem to f on [x1 , x2 ].
Problem 5 : Let a > 0 and f : [−a, a] → R be continuous. Suppose f (x) exists and f (x) ≤ 1 for
all x ∈ (−a, a). If f (a) = a and f (−a) = −a, then show that f (x) = x for every x ∈ (−a, a).
Solution : Let g(x) = f (x)−x on [−a, a]. Note that g (x) ≤ 0 on (−a, a). Therefore, g is decreasing.
Since g(a) = g(−a) = 0, we have g = 0.
This problem can also be solved by applying the MVT for g on [−a, x] and [x, a].](https://image.slidesharecdn.com/rollestheoremmeanvaluetheorem-121127120402-phpapp02/85/Rolle-s-theorem-mean-value-theorem-2-320.jpg)
Rolle's theorem, mean value theorem
- 1. 1 Lecture 6 : Rolle’s Theorem, Mean Value Theorem The reader must be familiar with the classical maxima and minima problems from calculus. For example, the graph of a differentiable function has a horizontal tangent at a maximum or minimum point. This is not quite accurate as we will see. Definition : Let f : I → R, I an interval. A point x0 ∈ I is a local maximum of f if there is a δ > 0 such that f (x) ≤ f (x0 ) whenever x ∈ I ∩ (x0 − δ, x0 + δ). Similarly, we can define local minimum. Theorem 6.1 : Suppose f : [a, b] → R and suppose f has either a local maximum or a local minimum at x0 ∈ (a, b). If f is differentiable at x0 then f (x0 ) = 0. Proof: Suppose f has a local maximum at x0 ∈ (a, b). For small (enough) h, f (x0 + h) ≤ f (x0 ). If h > 0 then f (x0 + h) − f (x0 ) ≤ 0. h Similarly, if h < 0, then f (x0 + h) − f (x0 ) ≥ 0. h By elementary properties of the limit, it follows that f (x0 ) = 0. We remark that the previous theorem is not valid if x0 is a or b. For example, if we consider the function f : [0, 1] → R such that f (x) = x, then f has maximum at 1 but f (x) = 1 for all x ∈ [0, 1]. The following theorem is known as Rolle’s theorem which is an application of the previous theorem. Theorem 6.2 : Let f be continuous on [a, b], a < b, and differentiable on (a, b). Suppose f (a) = f (b). Then there exists c such that c ∈ (a, b) and f (c) = 0. Proof: If f is constant on [a, b] then f (c) = 0 for all c ∈ [a, b]. Suppose there exists x ∈ (a, b) such that f (x) > f (a). (A similar argument can be given if f (x) < f (a)). Then there exists c ∈ (a, b) such that f (c) is a maximum. Hence by the previous theorem, we have f (c) = 0. Problem 1 : Show that the equation x13 + 7x3 − 5 = 0 has exactly one (real) root. Solution : Let f (x) = x13 + 7x3 − 5. Then f (0) < 0 and f (1) > 0. By the IVP there is at least one positive root of f (x) = 0. If there are two distinct positive roots, then by Rolle’s theorem there is some x0 > 0 such that f (x0 ) = 0 which is not true. Moreover, observe that f (x) < 0 for x < 0. Problem 2 : Let f and g be functions, continuous on [a, b], differentiable on (a, b) and let f (a) = f (b) = 0. Prove that there is a point c ∈ (a, b) such that g (c)f (c) + f (c) = 0. Solution : Define h(x) = f (x)eg(x) . Here, h(x) is continuous on [a, b] and differentiable on (a, b). Since h(a) = h(b) = 0, by Rolle’s theorem, there exists c ∈ (a, b) such that h (c) = 0. Since h (x) = [f (x)+g (x)f (x)]eg(x) and eα = 0 for any α ∈ R, we see that f (c)+g (c)f (c) = 0. A geometric interpretation of the above theorem can be given as follows. If the values of a differentiable function f at the end points a and b are equal then somewhere between a and b there is a horizontal tangent. It is natural to ask the following question. If the value of f at the end points a and b are not the same, is it true that there is some c ∈ [a, b] such that the tangent line at c is parallel to the line connecting the endpoints of the curve? The answer is yes and this is essentially the Mean Value Theorem.
- 2. 2 Theorem 6.3 : (Mean Value Theorem) Let f be continuous on [a, b] and differentiable on (a, b). Then there exists c ∈ (a, b) such that f (b) − f (a) = f (c)(b − a). Proof: Let f (b) − f (a) g(x) = f (x) − (x − a). b−a Then g(a) = g(b) = f (a). The result follows by applying Rolle’s Theorem to g. The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f . For example, if we have a property of f and we want to see the effect of this property on f , we usually try to apply the mean value theorem. Let us see some examples. Example 1 : Let f : [a, b] → R be differentiable. Then f is constant if and only if f (x) = 0 for every x ∈ [a, b]. Proof : Suppose that f is constant, then from the definition of f (x) it is immediate that f (x) = 0 for every x ∈ [a, b]. To prove the converse, let a < x ≤ b. By the mean value theorem there exists c ∈ (a, x) such that f (x)−f (a) = f (c)(x−a). Since f (c) = 0, we conclude that f (x) = f (a), that is f is constant. (If we try to prove the converse directly from the definition of f (x) we will be in trouble.) Example 2 : Suppose f is continuous on [a, b] and differentiable on (a, b). (i) If f (x) = 0 for all x ∈ (a, b), then f is one-one (i.e, f (x) = f (y) whenever x = y). (ii) If f (x) ≥ 0 (resp. f (x) > 0) for all x ∈ (a, b) then f is increasing (resp. strictly increasing) on [a, b]. (We have a similar result for decreasing functions.) Proof : Apply the mean value theorem as we did in the previous example. (Note that f can be one-one but f can be 0 at some point, for example take f (x) = x3 and x = 0.) Problem 3 : Use the mean value theorem to prove that | sinx − siny | ≤ | x − y | for all x, y ∈ R. Solution : Let x, y ∈ R. By the mean value theorem sinx − siny = cosc (x − y) for some c between x and y. Hence | sinx − siny | ≤ | x − y |. Problem 4 : Let f be twice differentiable on [0, 2]. Show that if f (0) = 0, f (1) = 2 and f (2) = 4, then there is x0 ∈ (0, 2) such that f (x0 ) = 0. Solution : By the mean value theorem there exist x1 ∈ (0, 1) and x2 ∈ (1, 2) such that f (x1 ) = f (1) − f (0) = 2 and f (x2 ) = f (2) − f (1) = 2. Apply Rolle’s theorem to f on [x1 , x2 ]. Problem 5 : Let a > 0 and f : [−a, a] → R be continuous. Suppose f (x) exists and f (x) ≤ 1 for all x ∈ (−a, a). If f (a) = a and f (−a) = −a, then show that f (x) = x for every x ∈ (−a, a). Solution : Let g(x) = f (x)−x on [−a, a]. Note that g (x) ≤ 0 on (−a, a). Therefore, g is decreasing. Since g(a) = g(−a) = 0, we have g = 0. This problem can also be solved by applying the MVT for g on [−a, x] and [x, a].