Question

help plz

Answer 1

\[\frac{ x-1 }{ 4 }\] \[if 0\le x \le 1/2\]


\[e ^{-x ^{2}}\] \[if 1/2 < x \le\]

Answer 2

second one is from
\[\Large \frac{1}{2} \le x \le \infty\] ?

Answer 3

Given the function

Answer 4

\[if 1/2 < x \le 1 \]

sorry the second is from

Answer 5

\[f(x)=\left\{\begin{matrix}\frac{x-1}{4} ,& 0\le x\le \frac{1}{2} \\ e^{-x^2}, & \frac{1}{2}<x\le 1\end{matrix}\right.\]

Answer 6

yes @Zarkon

Answer 7

and the directions for the problem are ...

Answer 8

given the funtion

Answer 9

i think to find continuity ???

Answer 10

am i right Muskan ?

Answer 11

if its to check for continuity at 1/2 then
left hand limit is
\[\Large \lim_{x \rightarrow \frac{1}{2}} \frac{x-1}{4}=\frac{\frac{1}{2}-1}{4}=\frac{-1}{8}\]
right hand limit limit is
\[\Large \lim_{x \rightarrow \frac{1}{2}} e^{-x^2}\]
\[\Large =e^{-(1/2)^2}=e^{-1/4}=0.778\]
since left and right hand limits are not equal so function is not continuous at x=1/2

Answer 12

Consider the function
\[f(x) = \frac{ x }{ \left| x \right| }\]
determines its domain. draws its graphics and reasonable if you can assign a value f (0) for the function is continuous at every IR.

Answer 13

@hartnn

Answer 14

for x/|x|

if x>0 then |x|=x and x/|x|=1
if x<0 then |x|=-x and x/|x|=-1

Answer 15

ok