The problem has two parts. Part (a) asks to determine the truthfulness of two statements and provide explanations or counterexamples. (i) If a function $f$ 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 exists $c \in (a, b)$ such that $f'(c) = 0$. Part (b) asks to find the values of all six hyperbolic functions at $x$, given that $\text{sech}(2x) = \frac{8}{17}$ and $x < 0$.

1. Problem Description

The problem has two parts.

Part (a) asks to determine the truthfulness of two statements and provide explanations or counterexamples.

(i) If a function $f$ 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 exists $c \in (a, b)$ such that $f'(c) = 0$.

Part (b) asks to find the values of all six hyperbolic functions at $x$, given that $\text{sech}(2x) = \frac{8}{17}$ and $x < 0$.

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 $f(x) = x^3 - 3x$ defined on the interval $I = [-2, 2]$. The derivative is $f'(x) = 3x^2 - 3$. Setting $f'(x) = 0$, we find $x = \pm 1$. So, $x=1$ and $x=-1$ are critical points.

$f''(x) = 6x$.

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

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

However, the global minimum occurs at $x=-2$ with $f(-2) = -8 + 6 = -2$, and the global maximum occurs at $x=2$ with $f(2) = 8 - 6 = 2$. While $f(1)$ is a local minimum, $f(1) = -2$ is not a global minimum. Similarly, while $f(-1)$ is a local maximum, $f(-1) = 2$ is not a global maximum. The global maximum and minimum occur at $x=2$ and $x=-2$ respectively. Therefore, $f(1)$ and $f(-1)$ are not global extreme values.

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

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

(b)

Given $\text{sech}(2x) = \frac{8}{17}$ and $x < 0$. We need to find the values of $\sinh x$, $\cosh x$, $\tanh x$, $\text{cosech } x$, $\text{sech } x$, and $\coth x$.

Since $\text{sech}(2x) = \frac{1}{\cosh(2x)}$, we have $\cosh(2x) = \frac{17}{8}$.

We know that

$\cosh(2x) = 2 \cosh^2(x) - 1$, so $2 \cosh^2(x) - 1 = \frac{17}{8}$.

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

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

Since $\cosh(x) > 0$ for all $x$, we have $\cosh(x) = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

Now, we know $\cosh^2(x) - \sinh^2(x) = 1$.

So, $\sinh^2(x) = \cosh^2(x) - 1 = \left(\frac{5}{4}\right)^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}$.

Since $x < 0$, $\sinh(x) < 0$, so $\sinh(x) = -\sqrt{\frac{9}{16}} = -\frac{3}{4}$.

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

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

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

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

3. Final Answer

(a)

(i) FALSE. Counterexample: $f(x) = x^3 - 3x$ on $[-2, 2]$.

(ii) TRUE. Rolle's Theorem.

(b)

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

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

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

$\text{cosech } x = -\frac{4}{3}$

$\text{sech } x = \frac{4}{5}$

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