Tag: Intermediate-Value Theorem

Prove the intermediate value theorem for derivatives

by RoRi

September 28, 2015

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

Assume $f$ is differentiable on an open interval $I$ . Let $a < b$ be two points in $I$ . Then, the derivative $f'$ takes every value between $f'(a)$ and $f'(b)$ somewhere in $(a,b)$ .

Define a function
$g(x) = \frac{f(x) - f(a)}{x-a} \qquad \text{if } x \neq a, \qquad g(a) = f'(a).$

Prove that $g$ takes every value between $f'(a)$ and $f'(b)$ in the interval $(a,b)$ . Then, use the mean-value theorem for derivatives to show $f'$ takes all values between $f'(a)$ and $g(b)$ somewhere in the interval $(a,b)$ .
Define a function
$h(x) = \frac{f(x) - f(b)}{x-b}, \qquad \text{if } x \neq b, \qquad h(b) = f'(b).$

Show that the derivative $f'$ takes on all values between $f'(b)$ and $h(a)$ in the interval $(a,b)$ . Conclude that the statement of the intermediate-value theorem is true.

Proof. First, since $f$ is differentiable everywhere on the interval $I$ , we know $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ . Thus, if $x \in [a,b]$ and $x \neq a$ then $g$ is continuous at $x$ since it is the quotient of continuous functions and the denominator is nonzero. If $x = a$ then
$\lim_{x \to a} g(x) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} = f'(a) = g(a);$

hence, $g$ is continuous at $x = a$ as well. Therefore, $g$ is continuous on the closed interval $[a,b]$ . So, by the intermediate value theorem for continuous functions we know $g$ takes on every value between $g(a)$ and $g(b)$ somewhere on the interval $(a,b)$ . Since $g(a) = f'(a)$ this means $g$ takes on every value between $f'(a)$ and $g(b)$ somewhere on the interval $(a,b)$ .
By the mean-value theorem for derivatives, we then know there exists some $c \in (a,b)$ such that

$\begin{align*} &&f(x) - f(a) &= f'(c)(x-a) \\ \implies && f'(c) &= \frac{f(x) - f(a)}{x-a} \\ \implies && f'(c) &= g(x) \end{align*}$

for some $c \in (a,x)$ . Since $x < b$ , we then conclude there is some $c \in (a,b)$ such that $f'(c) = g(x)$ for any $x$ . Since $g$ takes on every value between $f'(a)$ and $g(b)$ , so does $f'$ .
Proof. This is very similar to part (a). By the same argument we have the function $h$ is continuous on $[a,b]$ ; thus, $h$ takes on every value between $h(a)$ and $f'(b)$ by the intermediate value theorem for continuous functions. Then, by the mean value theorem, we know there exists a $c \in (x, b)$ such that
$f'(c) = \frac{f(x) - f(b)}{x-b} = h(x) \qquad \text{for some } c \in (x,b).$

Thus, $f'$ takes on every value between $f'(b)$ and $h(a)$ . Since $h(a) = g(b)$ ; $f'$ takes on every value between $f'(a)$ and $f'(b). \qquad \blacksquare$

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

by RoRi

September 28, 2015

Consider a function $f$ which is continuous everywhere on an interval $[a,b]$ and has a second derivative $f''$ everywhere on the open interval $(a,b)$ . Assume the chord joining two points $(a, f(a))$ and $(b, f(b))$ on the graph of the function intersects the graph of the function at a point $(c, f(c))$ with $c \in (a,b)$ . Prove there exists a point $t \in (a,b)$ such that $f''(t) = 0$ .

Proof. Let $g(x)$ be the equation of the line joining $(a, f(a))$ and $(b, f(b))$ . Then define a function

$h(x) = f(x) - g(x).$

Since $f$ and $g$ intersect at the values $a,b$ , and $c$ this means

$f(a) = g(a), \qquad f(b) = g(b), \qquad f(c) = g(c).$

By our definition of $h$ then we have

$h(a) = h(b) = h(c) = 0.$

Further, since $f'$ and $g'$ are continuous and differentiable on $(a,b)$ , we apply Rolle’s theorem twice: first on the interval $[a,c]$ and then on the interval $[c,b]$ . These two applications of Rolle’s theorem tells us there exist points $c_1$ and $c_2$ such that

$h'(c_1) = h'(c_2) = 0, \qquad \text{with } a < c_1 < c < c_2 < b.$

Then, we apply Rolle’s theorem for a third time, this time to the function $h'$ on the interval $[c_1, c_2]$ to conclude that there exists a $t \in (c_1, c_2)$ such that $h''(t) = 0$ . Then, since $t \in (c_1, c_2)$ we know $t \in (a,b)$ (since $a < c_1 < c_2 < b$ ). So,

$\begin{align*} h(x) = f(x) - g(x) && \implies && h'(x) &= f'(x) - g'(x) \\ && \implies && h''(x) &= f''(x) &(g''(x) = 0 \text{ since } g \text{ is linear}). \end{align*}$

Thus, $f''(t) = 0$ for some $t \in (a,b). \qquad \blacksquare$

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

by RoRi

September 27, 2015

Consider the equation

$x^2 = x \sin x + \cos x.$

Show that there are two values of $x \in \mathbb{R}$ such that the equation is satisfied.

Proof. Let $f(x) = x^2 - x \sin x - \cos x$ . (We want then to find the zeros of this function since these will be the points that $x^2 = x \sin + \cos x$ .) Then,

$f'(x) = 2x - \sin x - x \cos x + \sin x = x(2 - \cos x).$

Since $2 - \cos x \neq 0$ for any $x$ (since $\cos x \leq 1$ ), we have

$f'(x) = 0 \quad \iff \quad x = 0.$

Then, $f$ is continuous and differentiable everywhere, so we may apply Rolle’s theorem on any interval. So, by Rolle’s theorem we know $f$ has at most two zeros (if there were three or more, say $x_1, x_2,$ and $x_3$ , then there must be distinct numbers $c_1$ and $c_2$ with $x_1 < c_1 < x_2 < c_2 < x_3$ such that $f'(c_1) = f'(c_2) = 0$ , but we know there is only one $c$ such that $f'(c) = 0$ ).
Furthermore, $f$ has at least two zeros since $f(-\pi) = \pi^2 + 1 > 0$ , $f(0) = -1 < 0$ , and $f(\pi) = \pi^2 + 1 > 0$ . Thus, by Bolzano’s theorem there are zeros between each of these points. We have that the number of zeros of $f$ is at most two and at least 2. Hence, the number of zeros must be exactly two $. \qquad \blacksquare$

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

by RoRi

September 26, 2015

Prove, using Rolle’s theorem, that for any value of $b$ there is at most one $x \in [-1,1]$ such that

$x^3 - 3x + b = 0.$

Proof. The argument is by contradiction. Suppose there are two or more points in $[-1,1]$ for which $x^3 - 3x + b = 0$ . Let $x_1, x_2$ be two such points with $x_1 < x_2$ . Since $x_1, x_2$ are both in $[-1,1]$ we have

$-1 \leq x_1 < x_2 \leq 1 \qquad \text{and} \qquad f(x_1) = f(x_2) = 0.$

Since $f(x) = x^3 - 3x + b$ is continuous on $[-1,1]$ and differentiable on $(-1,1)$ (all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval $[x_1, x_2]$ to conclude that there is some $c \in (x_1, x_2)$ such that

$f'(c) = 0 \quad \implies \quad 3c^2 - 3 = 0 \quad \implies \quad c = \pm 1.$

But this contradicts that $c \in (-1,1)$ (since $c \in (x_1, x_2)$ and $-1 \leq x_1 < x_2 \leq 1$ ). Hence, there can be at most one point $x \in [-1,1]$ such that $f(x) = 0. \qquad \blacksquare$

Prove connection between power mean and arithmetic mean of a function

by RoRi

September 6, 2015

Let $f$ be a continuous, strictly monotonic function on $\mathbb{R}_{>0}$ with inverse $g$ , and let $a_1 < a_2 < \cdots < a_n$ be given positive real numbers. Then define,

$M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right).$

This $M_f$ is called the mean of $a_1, \ldots, a_n$ with respect to $f$ . (When $f(x) = x^p$ for $p \neq 0$ , this coincides with the $p$ th power mean from this exercise).

Show that

$f(M_f) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Proof. Since $g$ is the inverse of $f$ we know $f(g(x)) = x$ for all $x$ in the range of $f$ , i.e., for all $x$ such that there is some $c \in \mathbb{R}_{>0}$ such that $f(c) = x$ .

By the definition of $M_f$ then, we have that

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right).$

So, if $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ then we are done. Since $g$ is the inverse of $f$ it’s domain is equal to the range of $f$ . We show that this value is in the range of $f$ using the intermediate value theorem.

Without loss of generality, assume $f$ is strictly increasing (the alternative assumption, that $f$ is strictly decreasing will produce an almost identical argument). Then, since $a_1 < a_2 < \cdots < a_n$ are all positive real numbers we have $f(a_1) < f(a_2) < \cdots < f(a_n)$ . (Here if we’d assumed that $f$ was strictly decreasing the roles inequalities would be reversed.) Then we have,

$f(a_1) = \frac{1}{n} \sum_{i=1}^n f(a_1) < \frac{1}{n} \sum_{i=1}^n f(a_i) < \frac{1}{n} \sum_{i=1}^n f(a_n) = f(a_n).$

Hence, by the intermediate value theorem, since

$\frac{1}{n} \sum_{i=1}^n f(a_i) \in [f(a_1), f(a_n)]$

there must be some $c \in \mathbb{R}_{>0}$ such that

$f(c) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Thus, $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ , so

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right) = \frac{1}{n} \sum_{i=1}^n f(a_i). \qquad \blacksquare$

Prove a function with given properties has a fixed point

by RoRi

September 5, 2015

Given a real function $f$ continuous on $[a,b]$ , with $f(a) \leq a$ and $f(b) \geq b$ . Prove there is some $c \in [a,b]$ such that $f(c) = c$ (i.e., $f$ has a fixed point in $[a,b]$ ).

Proof. If $f(a) = a$ or $f(b) = b$ then we are done since these would be fixed points.

Assume then that $f(a) < a$ and $f(b) > b$ and let $g(x) = f(x) - x$ . We know that $g$ is continuous on $[a,b]$ and

$g(a) = f(a) - a < 0, \qquad \text{and} \qquad g(b) = f(b) - b > 0.$

Thus, by Bolzano’s theorem, there is some $c \in [a,b]$ such that $g(c) = 0$ ; hence, $f(c) - c = 0$ or $f(c) = c. \qquad \blacksquare$

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

by RoRi

September 5, 2015

Let $f$ be a continuous, real function on the interval $[0,1]$ . Assume

$0 \leq f(x) \leq 1 \qquad \text{for all } x \in [0,1].$

Prove that there exists a real number $c \in [0,1]$ such that $f(c) = c$ .

Proof. Let $g(x) = f(x) - x$ . Then $g$ is continuous on $[0,1]$ since it is the difference of functions which are continuous on $[0,1]$ .

Then, $g(0) = f(0)$ and $g(1)= f(1) - 1$ . If $f(0) = 0$ , then $c = 0$ and we are done. Similarly, if $f(1) = 1$ , then $c = 1$ and we are done as well.

Assume then that $0 < f(0) \leq 1$ , and $0 \leq f(1) < 1$ . Then

$\begin{align*} f(0) > 0 \quad \implies \quad g(0) &>0 \\ f(1) < 1 \quad \implies \quad g(1) &= f(1) - 1 < 0. \end{align*}$

Hence, applying Bolzano’s theorem to $g$ , there is some $c \in [0,1]$ such that $g(c) = 0$ . This implies $f(c) - c = 0$ , or $f(c) = c. \qquad \blacksquare$

Explain why tan x does not violate Bolzano’s theorem

by RoRi

September 5, 2015

Letting $f(x) = \tan x$ we observe that

$f \left( \frac{\pi}{4} \right) = \tan \frac{\pi}{4} = 1, \quad \text{and} \quad f \left( \frac{3 \pi}{4} \right) = \tan \frac{3 \pi}{4} = -1.$

However, there is no $x \in \left[ \frac{\pi}{4}, \frac{3 \pi}{4} \right]$ such that $f(x) =0$ . Why is this not a counterexample to Bolzano’s theorem?

This does not contradict Bolzano’s theorem since $\tan x$ is not continuous on the interval $\left[ \frac{\pi}{4}, \frac{3 \pi}{4} \right]$ since it is not defined at $\frac{\pi}{2}$ .

Prove there is exactly one negative solution to an equation

by RoRi

September 5, 2015

Show there is exactly one $b < 0$ such that $b^n = a$ for $a < 0$ and $n$ an odd, positive integer.

Proof. Let $f(x) = x^n$ , and let $c < -1$ with $c < a < 0$ . Then, $c^n < c$ (for odd $n$ ) so $c^n < c < a < 0$ . Since $f(0) = 0$ and $f(c) = c^n$ , by the Intermediate Value Theorem, we know $f(x)$ takes every value between $c^n$ and 0 for some $x \in [c,0]$ . Thus, we know there exists $b \in [c,0]$ such that $f(b) = a$ (since $c^n < a < 0$ ). This implies $b^n = a$ for some $b \in [c,0]$ .

We know this solution is unique since $f$ is strictly increasing on the whole real line for odd $n. \qquad \blacksquare$

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

by RoRi

September 4, 2015

Define a polynomial

$f(x) = \sum_{k=0}^n c_k x^k,$

such that $c_0$ and $c_n$ have opposite signs. Prove there is some $x > 0$ such that $f(x) = 0$ .

Proof. Since

$f(0) = \sum_{k=0}^n c_k 0^k = c_0$

we see that $f(0)$ has the same sign as $c_0$ .
Next,

$f(x) = c_n x^n + c_{n-1}x^{n-1} + \cdots + c_1 x + c_0 = c_n x^n \left( 1+ \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right).$

Then we claim that for sufficiently large $x$ ,

$1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} > 0;$

hence, $f(x)$ will have the same sign as $c_n$ for sufficiently large $x$ , since $x > 0$ and so

$x^n \left( 1 + \frac{c_{n-1}}{c_n} \frac{1}{x} + \cdots + \frac{c_0}{c_n} \frac{1}{x^n} \right) > 0.$

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

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

$\left( 1 + \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} + \cdots + \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right) \geq 1 - \left( \left| \frac{c_{n-1}}{c_n} \cdot \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \cdot \frac{1}{x^n} \right| \right).$

This is true since for each term in the sum

$\left| \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}} \right| \geq \frac{c_i}{c_n} \cdot \frac{1}{x^{n-i}}, \qquad \text{for all } 1 \leq i \leq n-1.$

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

$x > \max \left\{ 1, \left| \frac{c_n}{n \cdot c_a} \right| \right\}, \qquad \text{where} \qquad |c_a| = \max\{ |c_0|, |c_1|, \ldots, |c_n|\}.$

Since $x > 1$ , we know $\frac{1}{x} > \frac{1}{x^n}$ for all $n \in \mathbb{Z}_{>0}$ , so

$\begin{align*} 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x^n} \right| \right) &> 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{1}{x} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{1}{x} \right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{c_n} \frac{c_n}{n c_a} \right| + \cdots + \left| \frac{c_0}{c_n} \frac{c_n}{n c_a}\right| \right) \\ &= 1 - \left( \left| \frac{c_{n-1}}{n c_a} \right| + \cdots + \left| \frac{c_0}{nc_a} \right| \right) \\ &> 1 - \left( \left| \frac{c_a}{nc_a} \right| + \cdots + \left| \frac{c_a}{nc_a} \right| \right)\\ &= 1 - \left( \underbrace{\frac{1}{n} + \cdots + \frac{1}{n}}_{n-1\text{ terms}} \right) \\ &= 1 - \left( \frac{n-1}{n} \right) \\ &= \frac{1}{n} > 0. \end{align*}$

This proves our claim, and so $f(x)$ has the same sign as $c_n$ for sufficiently large $x$ . Hence, $f(0)$ and $f(x)$ have different signs (since $c_0$ and $c_n$ had different signs by assumption); thus, there is some $c>0$ such that $f(c) = 0. \qquad \blacksquare$