The problem has two parts. a) We are given two statements about functions and intervals, and we need to determine if they are true or false. If true, we explain why. If false, we provide a counterexample. (i) If $f$ has a local extreme value at $a$, then $f$ has a global/absolute extreme value at $a$. (ii) If $a < b$ and $f(a) = f(b)$, then there is $c \in (a, b)$ such that $f'(c) = 0$. b) Given that $\mathrm{sech}(2x) = \frac{8}{17}$ and $x < 0$, we need to find the values of the six hyperbolic functions: $\sinh x$, $\cosh x$, $\tanh x$, $\mathrm{cosech}\, x$, $\mathrm{sech}\, x$, and $\coth x$.

1. Problem Description

The problem has two parts.

a) We are given two statements about functions and intervals, and we need to determine if they are true or false. If true, we explain why. If false, we provide a counterexample.

(i) If $f$ has a local extreme value at $a$, then $f$ has a global/absolute extreme value at $a$.

(ii) If $a < b$ and $f(a) = f(b)$, then there is $c \in (a, b)$ such that $f'(c) = 0$.

b) Given that $\mathrm{sech}(2x) = \frac{8}{17}$ and $x < 0$, we need to find the values of the six hyperbolic functions: $\sinh x$, $\cosh x$, $\tanh x$, $\mathrm{cosech}\, x$, $\mathrm{sech}\, x$, and $\coth x$.

2. Solution Steps

a)

(i) The statement "If $f$ has a local extreme value at $a$, then $f$ has a global/absolute extreme value at $a$" is FALSE.

Counterexample: Consider the function $f(x) = x^3 - 3x$ on the interval $[-2, 2]$. The derivative is $f'(x) = 3x^2 - 3$. Setting $f'(x) = 0$, we have $3x^2 - 3 = 0$, so $x^2 = 1$, and $x = \pm 1$. Thus, $x=1$ and $x=-1$ are critical points.

$f''(x) = 6x$. $f''(1) = 6 > 0$, so $f(1) = 1 - 3 = -2$ is a local minimum. $f''(-1) = -6 < 0$, so $f(-1) = -1 + 3 = 2$ is a local maximum.

However, $f(-2) = -8 + 6 = -2$ and $f(2) = 8 - 6 = 2$. Thus, $f(1)$ and $f(-1)$ are not global extrema because $f(-2) = -2$ and $f(2) = 2$.

(ii) The statement "If $a < b$ and $f(a) = f(b)$, then there is $c \in (a, b)$ such that $f'(c) = 0$" is TRUE.

This is Rolle's Theorem. If a function $f$ is continuous on the closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that $f'(c) = 0$.

b)

Given $\mathrm{sech}(2x) = \frac{8}{17}$ and $x < 0$.

We have $\mathrm{cosh}(2x) = \frac{1}{\mathrm{sech}(2x)} = \frac{17}{8}$.

Using the identity $\mathrm{cosh}(2x) = 2 \mathrm{cosh}^2(x) - 1$, we get $2 \mathrm{cosh}^2(x) - 1 = \frac{17}{8}$.

$2 \mathrm{cosh}^2(x) = \frac{17}{8} + 1 = \frac{25}{8}$.

$\mathrm{cosh}^2(x) = \frac{25}{16}$.

$\mathrm{cosh}(x) = \pm \frac{5}{4}$.

Since $\mathrm{cosh}(x)$ is always positive, we have $\mathrm{cosh}(x) = \frac{5}{4}$.

Using the identity $\mathrm{sinh}^2(x) = \mathrm{cosh}^2(x) - 1$, we get $\mathrm{sinh}^2(x) = (\frac{5}{4})^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}$.

$\mathrm{sinh}(x) = \pm \frac{3}{4}$. Since $x < 0$, $\sinh(x) < 0$, so $\sinh(x) = -\frac{3}{4}$.

$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{-3/4}{5/4} = -\frac{3}{5}$.

$\mathrm{cosech}(x) = \frac{1}{\sinh(x)} = \frac{1}{-3/4} = -\frac{4}{3}$.

$\mathrm{sech}(x) = \frac{1}{\cosh(x)} = \frac{1}{5/4} = \frac{4}{5}$.

$\coth(x) = \frac{1}{\tanh(x)} = \frac{1}{-3/5} = -\frac{5}{3}$.

3. Final Answer

a)

(i) False. Example: $f(x) = x^3 - 3x$.

(ii) True. This is Rolle's Theorem.

b)

$\sinh x = -\frac{3}{4}$

$\cosh x = \frac{5}{4}$

$\tanh x = -\frac{3}{5}$

$\mathrm{cosech}\, x = -\frac{4}{3}$

$\mathrm{sech}\, x = \frac{4}{5}$

$\coth x = -\frac{5}{3}$