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# Prove the intermediate value theorem for derivatives

Consider the following statement of the intermediate value theorem for derivatives:

Assume is differentiable on an open interval . Let be two points in . Then, the derivative takes every value between and somewhere in .

1. Define a function

Prove that takes every value between and in the interval . Then, use the mean-value theorem for derivatives to show takes all values between and somewhere in the interval .

2. Define a function

Show that the derivative takes on all values between and in the interval . Conclude that the statement of the intermediate-value theorem is true.

1. Proof. First, since is differentiable everywhere on the interval , we know is continuous on and differentiable on . Thus, if and then is continuous at since it is the quotient of continuous functions and the denominator is nonzero. If then

hence, is continuous at as well. Therefore, is continuous on the closed interval . So, by the intermediate value theorem for continuous functions we know takes on every value between and somewhere on the interval . Since this means takes on every value between and somewhere on the interval .
By the mean-value theorem for derivatives, we then know there exists some such that

for some . Since , we then conclude there is some such that for any . Since takes on every value between and , so does .

2. Proof. This is very similar to part (a). By the same argument we have the function is continuous on ; thus, takes on every value between and by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a such that

Thus, takes on every value between and . Since ; takes on every value between and

# Prove the second derivative of a function with a given property must have a zero

Consider a function which is continuous everywhere on an interval and has a second derivative everywhere on the open interval . Assume the chord joining two points and on the graph of the function intersects the graph of the function at a point with . Prove there exists a point such that .

Proof. Let be the equation of the line joining and . Then define a function

Since and intersect at the values , and this means

By our definition of then we have

Further, since and are continuous and differentiable on , we apply Rolle’s theorem twice: first on the interval and then on the interval . These two applications of Rolle’s theorem tells us there exist points and such that

Then, we apply Rolle’s theorem for a third time, this time to the function on the interval to conclude that there exists a such that . Then, since we know (since ). So,

Thus, for some

# Show that x^2 = x*sin x + cos x for exactly two real numbers x

Consider the equation

Show that there are two values of such that the equation is satisfied.

Proof. Let . (We want then to find the zeros of this function since these will be the points that .) Then,

Since for any (since ), we have

Then, is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know has at most two zeros (if there were three or more, say and , then there must be distinct numbers and with such that , but we know there is only one such that ).
Furthermore, has at least two zeros since , , and . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of is at most two and at least 2. Hence, the number of zeros must be exactly two

# Prove there is only one zero of x3 – 3x + b in [-1,1]

Prove, using Rolle’s theorem, that for any value of there is at most one such that

Proof. The argument is by contradiction. Suppose there are two or more points in for which . Let be two such points with . Since are both in we have

Since is continuous on and differentiable on (all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval to conclude that there is some such that

But this contradicts that (since and ). Hence, there can be at most one point such that

# Prove connection between power mean and arithmetic mean of a function

Let be a continuous, strictly monotonic function on with inverse , and let be given positive real numbers. Then define,

This is called the mean of with respect to . (When for , this coincides with the th power mean from this exercise).

Show that

Proof. Since is the inverse of we know for all in the range of , i.e., for all such that there is some such that .

By the definition of then, we have that

So, if is in the domain of then we are done. Since is the inverse of it’s domain is equal to the range of . We show that this value is in the range of using the intermediate value theorem.

Without loss of generality, assume is strictly increasing (the alternative assumption, that is strictly decreasing will produce an almost identical argument). Then, since are all positive real numbers we have . (Here if we’d assumed that was strictly decreasing the roles inequalities would be reversed.) Then we have,

Hence, by the intermediate value theorem, since

there must be some such that

Thus, is in the domain of , so

# Prove a function with given properties has a fixed point

Given a real function continuous on , with and . Prove there is some such that (i.e., has a fixed point in ).

Proof. If or then we are done since these would be fixed points.

Assume then that and and let . We know that is continuous on and

Thus, by Bolzano’s theorem, there is some such that ; hence, or

# Prove that a particular function has a fixed point on [0,1]

Let be a continuous, real function on the interval . Assume

Prove that there exists a real number such that .

Proof. Let . Then is continuous on since it is the difference of functions which are continuous on .

Then, and . If , then and we are done. Similarly, if , then and we are done as well.

Assume then that , and . Then

Hence, applying Bolzano’s theorem to , there is some such that . This implies , or

# Explain why tan x does not violate Bolzano’s theorem

Letting we observe that

However, there is no such that . Why is this not a counterexample to Bolzano’s theorem?

This does not contradict Bolzano’s theorem since is not continuous on the interval since it is not defined at .

# Prove there is exactly one negative solution to an equation

Show there is exactly one such that for and an odd, positive integer.

Proof. Let , and let with . Then, (for odd ) so . Since and , by the Intermediate Value Theorem, we know takes every value between and 0 for some . Thus, we know there exists such that (since ). This implies for some .

We know this solution is unique since is strictly increasing on the whole real line for odd

# Prove a polynomial with opposite signed first and last coefficients has at least one positive zero

Define a polynomial

such that and have opposite signs. Prove there is some such that .

Proof. Since

we see that has the same sign as .
Next,

Then we claim that for sufficiently large ,

hence, will have the same sign as for sufficiently large , since and so

(So, when we multiply a positive number by the result will have the same sign as .)

So, we need to show that the claimed term is indeed positive. First,

This is true since for each term in the sum

Now, since we are showing there is sufficiently large such that our claim is true, we let

Since , we know for all , so

This proves our claim, and so has the same sign as for sufficiently large . Hence, and have different signs (since and had different signs by assumption); thus, there is some such that