Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 13

=Chapter 13=

1
$$f(x) = \frac{1}{x} \,$$

$$f\left(\frac{3}{4}\right) = \,$$ $$\frac {1}{\frac{3}{4}}    =  \,$$ $$1 * \left(\frac {3}{4}\right)^{-1}= \frac{4}{3}      \,$$

$$f\left(-\frac{2}{3}\right) = \,$$ $$\frac {1}{-\frac{2}{3}} = \,$$ $$1 * \left( \frac{-2}{3} \right)^{-1} = \,$$ $$ \frac{3}{-2} \,$$

2
$$f(x) = \frac{1}{x^2 - 2} \,$$ $$(x \neq \pm \sqrt{2}) \,$$

$$f(5) = \frac{1}{5^2 - 2} = \frac{1}{25 - 2} = \frac{1}{23}\,$$

3
$$f(x) = \sqrt[3]{x} \,$$ is defined for all real numbers.

$$f(27) = \sqrt[3]{27} = 3 \,$$ because $$3^3 = 27. \,$$

4
$$ a) f(1) = 1 \,$$

$$ b) f(-3) = 3 \,$$

$$ c) f\left(-\frac{4}{3}\right) = \frac{4}{3} \,$$

5
$$ a) f\left(\frac{1}{2}\right) = 1/2 + |1/2| = 1 \,$$

$$ b) f(2) = 2 + |2| = 4 \,$$

$$ c) f(-4) = -4 + |-4| = -4 + 4 = 0\,$$

$$ d)f(-5) = -5 + |-5| = -5 + 5 = 0 \,$$

6
$$ f(x) = 2x + x^2 - 5 \,$$

$$a) f(1) = 2(1) + (1)^2 - 5 = -2 \,$$

$$b) f(-1) = 2(-1) + (-1)^2 - 5 = -6 \,$$

7
$$ f(x) = \sqrt[4]{x} \,$$

$$ x \geq 0 \,$$

$$ f(16) = 2, \,$$ because $$ 2^4 = 16 \,$$

8
$$ f(x) = -f(x) \,$$ is said to be an even function for all numbers x.

$$ f(x) = -f(-x) \,$$ is said to be an odd function for all x.

$$a)f(x) = x \,$$ is odd.

$$b)f(x) = x^2 \,$$ is even.

$$c)f(x) = x^3 \,$$ is odd.

$$d)f(x) = \frac{1}{x} \,$$ if $$x \neq 0 \,$$ and $$f(0) = 0 \,$$ is odd.

9
Let $$f_e(x) = \frac {f(x) + f(-x)}{2} \,$$ and $$f_o(x) = \frac {f(x) -f(-x)}{2}  \,$$ be the even and odd functions respectively.

Then $$f_e(x) + f_o(x) = \frac {f(x) + f(-x) + f(x) - f(-x)}{2} = f_e(x) \,$$

10
a) Odd

b) Even

c) Odd

d) Odd

e) Even

f) Even

g) Even

11
$$a)\,$$ $$f_a(x) = \frac{f(x) - f(-x)}{2} \,$$ $$ f_b(x) = \frac{f(x) - f(-x)}{2} \,$$

$$ f_a(x) + f_b(x) = \frac {f(x) - f(-x) + f(x) - f(-x)}{2} \,$$

It follows that $$ f_a(x) + f_b(x) = f(x) - f(-x) \,$$ (The odd function)

$$b)\,$$ $$f_a(x) = \frac{f(x) + f(-x)}{2} \,$$ $$ f_b(x) = \frac{f(x) + f(-x)}{2} \,$$

$$ f_a(x) + f_b(x) = \frac {f(x) + f(-x) + f(x) + f(-x)}{2} \,$$

It follows that $$ f_a(x) + f_b(x) = f(x) + f(-x) \,$$ (The even function)