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

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

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

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

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

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

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

${\displaystyle a)f(1)=1\,}$

${\displaystyle b)f(-3)=3\,}$

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

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

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

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

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

${\displaystyle f(x)=2x+x^{2}-5\,}$

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

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

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

${\displaystyle x\geq 0\,}$

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

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

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

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

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

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

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

Let ${\displaystyle f_{e}(x)={\frac {f(x)+f(-x)}{2}}\,}$ and ${\displaystyle f_{o}(x)={\frac {f(x)-f(-x)}{2}}\,}$ be the even and odd functions respectively.

Then ${\displaystyle f_{e}(x)+f_{o}(x)={\frac {f(x)+f(-x)+f(x)-f(-x)}{2}}=f_{e}(x)\,}$

a) Odd

b) Even

c) Odd

d) Odd

e) Even

f) Even

g) Even

${\displaystyle a)\,}$ ${\displaystyle f_{a}(x)={\frac {f(x)-f(-x)}{2}}\,}$ ${\displaystyle f_{b}(x)={\frac {f(x)-f(-x)}{2}}\,}$

${\displaystyle f_{a}(x)+f_{b}(x)={\frac {f(x)-f(-x)+f(x)-f(-x)}{2}}\,}$

It follows that ${\displaystyle f_{a}(x)+f_{b}(x)=f(x)-f(-x)\,}$ (The odd function)

${\displaystyle b)\,}$ ${\displaystyle f_{a}(x)={\frac {f(x)+f(-x)}{2}}\,}$ ${\displaystyle f_{b}(x)={\frac {f(x)+f(-x)}{2}}\,}$

${\displaystyle f_{a}(x)+f_{b}(x)={\frac {f(x)+f(-x)+f(x)+f(-x)}{2}}\,}$

It follows that ${\displaystyle f_{a}(x)+f_{b}(x)=f(x)+f(-x)\,}$ (The even function)