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Chapter 1: Q. 1 (page 134)

Values of transcendental functions: Without a calculator, find each of the function values that follow. For some values the answer may be undefined.

Short Answer

Expert verified

The value of each expression is calculated.

Step by step solution

01

Step 1. Given information

We have to find the value of the given functions.

02

Step 2. Values of the function.

When

$\begin{matrix} f (x) = \csc x \\ = \frac{1}{\sin 鈦� x} \\ F o r x = 蟺 \\ = \frac{1}{0} \\ = undefined \end{matrix}$

For $x = \frac{蟺}{2}$

$\begin{aligned} f (蟺 / 2) = \csc 鈦� (蟺 / 2) \\ = \frac{1}{\sin 鈦� (蟺 / 2)} \\ = \frac{1}{1} \\ = 1 \end{aligned}$

When $f (x) = \tan^{2} 鈦� x$

For $x = 蟺$

localid="1668092908461" $\begin{aligned} f (蟺) = \tan^{2} 鈦� 蟺 \\ = 0 \end{aligned}$

For $x = \frac{蟺}{2}$

$\begin{matrix} f (蟺 / 2) = \tan^{2} 鈦� (蟺 / 2) \\ = undefine \end{matrix}$

When $f (x) = \sin^{鈭� 1} 鈦� x$

For x=-1

$\begin{aligned} f (鈭� 1) = \sin^{鈭� 1} 鈦� (鈭� 1) \\ = \frac{鈭� 蟺}{2} \end{aligned}$

For $x = \frac{1}{2}$

$\begin{aligned} f (\frac{1}{2}) = \sin^{鈭� 1} 鈦� (\frac{1}{2}) \\ = \frac{蟺}{6} \end{aligned}$

When $f (x) = \tan^{鈭� 1} 鈦� \sqrt{x}$

For x=1

$\begin{aligned} f (1) = \tan^{鈭� 1} 鈦� \sqrt{1} \\ = \frac{蟺}{4} \end{aligned}$

For x=3

$\begin{aligned} f (3) = \tan^{鈭� 1} 鈦� \sqrt{3} \\ = \frac{蟺}{3} \end{aligned}$

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Most popular questions from this chapter

Sketch a labeled graph of a function that satisfies the hypothesis of the Extreme Value Theorem, and illustrate on your graph that the conclusion of the Extreme Value Theorem follows.

Each function in Exercises 9鈥�12 is discontinuous at some value x = c. Describe the type of discontinuity and any one-sided continuity at x = c, and sketch a possible graph of f.

$\lim_{x 鈫� 2^{-}} f (x) = 2, \lim_{x 鈫� 2^{+}} f (x) = 1, f (2) = 1 .$

Calculate each of the limits:

$\lim_{h 鈫� 0} \frac{{(- 1 + h)}^{3} - {(- 1)}^{3}}{h}$ .

$\lim_{x 鈫� 1} \frac{1 - \cos (x - 1)}{x}$

For each limit statement , use algebra to find 未 > 0 in terms of $蔚$ > 0 so that if 0 < |x 鈭� c| < 未, then | f(x) 鈭� L| < $蔚$ .

$\lim_{x 鈫� 1} (3 x + 8) = 11$

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