Solutions To Mathematics Textbooks/Basic Mathematics/Chapter 13

Chapter 13

Section 1

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)