Question

Graphical functions

Answer 1

\(\large \color{black}{\begin{align} &a.) \ f(x)=-f(x) \hspace{.33em}\\~\\
&b.) \ f(x)=f(-x) \hspace{.33em}\\~\\
&c.) \ \normalsize \text{neither even nor odd function} \hspace{.33em}\\~\\
&d.) \ f(x)\ \normalsize \text{doesn't exist at atleast one point of the domain.} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

it is not ? \[f(x)=cotan(x)\]
because of vertical retricemptotes ,and roots

Answer 3

i don't have any clue about the equation of the graph

Answer 4

vertical asymptote !
i had a mistake
if f(x)=cotan(x)
\[f(-x)=\cot(-x)=\frac{\cos(-x)}{\sin(-x)}=\frac{\cos(x)}{-\sin(x)}=-\cot(x) =-f(x)\]

Answer 5

it looks : the roots have a symmetry
I think they show roots ,nothing more
if they are part of figure , this is not a function

Answer 6

btw, choice a) f(x)=-f(x)
probably should read f(x) = - f(-x)

f(x)= -f(x) would mean y=-y which is only true for y=0

Answer 7

m confused is the answer d.)

Answer 8

domain of this function has not 1 point out R
in your picture ,\[D_f=\mathbb{R} -\left\{ 0,a,b \right\}\]

Answer 9

what is a and b

Answer 10

i'd say the lines just indicate zero crossing points and vertical asymptotes

Answer 11

From a textbook

this is the graph of y = cot x
the domain of the secant function is the set of all real numbers except those of the form 1/2 pi + n pi, and the function is continuous on its domain. Because with limit sec x = +/- angle, the graph of the secant function has the lines x = 1/2 pi + nr as vertical asymptotes. There is no intersection of the graph with the x axis because sec x is never zero. The secant function is even because