Exercises: Derivatives 3 – Serlo

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Rolle's theorem and the intermediate value theorem [Bearbeiten]

Exercise

Is there a such that has two distinct zeros in ?

Solution

Suppose there is a such that has two zeros . Since is continuous on and differentiable on , according to Rolle's theorem there is a with . But now there is for every : . So cannot have two zeros.

Exercise

Let be differentiable and continuous. Let further , and . Show that has at least one zero.

Proof

is continuous on and . Therefore, according to the intermediate value theorem, there is an with .

Further, is continuous on , differentiable on and . According to Rolle's theorem, there is hence a with . So has at least one zero (on ).

Mean-value theorem[Bearbeiten]

Exercise (An easy application of the mean-value theorem)

Let , . Show that there is a with ?

Solution (An easy application of the mean-value theorem)

is continuous on and differentiable on , as a composition of continuous and differentiable functions, respectively. Thus the mean value theorem is applicable. There is hence a with

Exercise (Useful inequality 1)

Show that: For all there is

Solution (Useful inequality 1)

Proof step:

Fall 1:

We define by . Then is continuous, and on differentiable. Thus the mean value theorem is applicable, and there exists a with

Now, because of ,

Fall 2:

Here we have , i.e., equality.

Fall 3:

Again, by the mean-value theorem there is a with

So there is for all

Proof step:

Here we show only the case :

We again define . Then is continuous, and on differentiable. According to the mean value theorem, there exists a with

For there is again equality, and for the statement follows analogously with the mean value theorem.

Hint

From the first inequality, the transition to for still allows for the inequality:

Exercise (Useful inequality 2)

Show that: For there is

Solution (Useful inequality 2)

Proof step:

Let . Then the sine function on is continuous and on it is differentiable. With the mean value theorem there is a with

But now there is for . Thus we have

Proof step:

Let . Then the tangent function on is continuous and on it is differentiable. With the mean value theorem there is a with

But now there is for . Thus we have

Hint

The inequality can be further extended to all :

Where equality only holds at .

Exercise (Implication of the mean value theorem)

Show by the mean value theorem that:

Let be continuous on and differentiable on . Furthermore, let and hold on . Then, there is on .

As an application: prove the following generalization of the Bernoulli inequality: For and all there is .

How to get to the proof? (Implication of the mean value theorem)

We present three different possible solutions: One with use of the mean value theorem, one via the monotony criterion and one via the fundamental theorem of calculus (missing). Within all three, we make use of the auxiliary function .

Proof (Implication of the mean value theorem)

We consider the auxiliary function

This function is continuous and differentiable on . Further there is

  1. for all

1st way: By the mean value theorem

By the mean value theorem, for all there is a with

for all

2nd way: By the monotony criterion

It follows from 1 that is monotonically increasing on (even on ).

for all for all

3rd way: By the fundamental theorem of calculus

By assumption, is integrable and because of the monotonicity of the integral there is for all :

But now, by the fundamental theorem of calculus

Concerning the application exercise: We define

and

Then and are continuous on and differentiable on with

and

Furthermore, there is . Since the exponential function is strictly monotonically increasing, there is for all :

With the proven statement we hence get for all :

Hint

If we even have and on , then we have on .

Hint

The generalized Bernoulli inequality can even be shown for all . Equality only holds in the case .

Exercise (mean valuer theorem for continuously differentiable functions)

Let be continuous, and continuously differentiable on . Show that there is a with , using the intermediate value theorem.

Solution (mean valuer theorem for continuously differentiable functions)

We consider any function with the given properties and the secant through the points and . The slope of the secant is given by the difference quotient . Next we look at the slope of the graph, i.e. the derivative values of the function on the interval .

Fall 1: The function graph is a straight line.

Then the derivative function is constant and consequently there is for all .

Fall 2: The function graph is not a straight line.

Then a must exist with or so that has no straight line as a function graph. It follows in turn that a exists with or , because otherwise the graph at can never take the function value . So in total there exist with . According to the intermediate value theorem, which is applicable here to the derivative function because it is continuous, there is now a with .

So in every m case there is a with .

Exercise (Unbounded derivative and uniform continuity)

Let be a differentiable function with . Prove that then is not a uniformly continuous function.

Proof (Unbounded derivative and uniform continuity)

Let be a differentiable function with . To prove that is not uniformly continuous, it must be shown that there is an , such that for all there are real numbers with and .

Choose . Now let be arbitrary. Because of there is an with for all with . According to the mean value theorem equation there is then for all with :

Now choose . There is by the above estimation and we get:

Thus is not uniformly continuous.

Exercise (Application of the second mean value theorem)

Let be differentiable. Further let for all . Show that then also

holds for all .

Proof (Application of the second mean value theorem)

Let be arbitrary with . Then and are continuous on and differentiable on according to the assumption. Then, with the second mean value theorem, there is a with

Since by assumption holds for all , we get .

From this we obtain

Lipschitz-continuity of functions[Bearbeiten]

Exercise (Lipschitz-continuity of functions)

Show by the mean value theorem (using the implication function about Lipschitz continuous functions), that the following functions are Lipschitz continuous. Determine in addition some suitable Lipschitz constants.

Solution (Lipschitz-continuity of functions)

Part 1: For all there is

So has a bounded derivative, and is therefore Lipschitz continuous. Further for all there is

Therefore, is an appropriate Lipschitz constant.

Part 2: Here, for all there is:

Thus, by boundedness of derivative, is Lipschitz-continuous, as well. Moreover, there is for all

Hence, is an appropriate Lipschitz constant here.

Part 3: finally there is for all :

So also is Lipschitz-continuous, and for all there is

Hence, is an appropriate Lipschitz constant.

Exercise (Lipschitz-continuity of functions)

Let be a continuous function that is differentiable on . Let be any two real numbers such that for all . Prove that for all the following estimate holds:

Proof (Lipschitz-continuity of functions)

Let be a continuous function with the properties of the problem. Let be arbitrary. If then and , so the above estimate is satisfied. Let therefore in the following .

Proof step:

Proof by contradiction: Let . Thus . It follows from the mean value theorem that there is an with . Thus is inconsistent with for all .

Proof step:

Proof by contradiction: Let . Thus . It follows from the mean value theorem that there is an with . Thus is inconsistent with for all .

Exercise (Local Lipschitz continuity)

Let be a continuously differentiable function, where is an open subset of . Prove that is locally Lipschitz continuous. That means, you need to prove that for all there exists an such that is Lipschitz- ontinuous on .

Proof (Local Lipschitz continuity)

Let be a continuously differentiable function, where is open. Let be arbitrary. Since is open, there is a such that . Choose . Then . Since is continuous, is a bounded function on .

Now we have already proved that differentiable functions with bounded derivatives are Lipschitz continuous. Thus is Lipschitz continuous on and thus also on .