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Prove the intermediate value theorem for derivatives

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).

  1. 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).

  2. 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.

  1. 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'.

  2. 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

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

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]

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

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

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

Prove there is exactly one negative solution to an equation

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

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.

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

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