Question

3 x (1/2)^x
Can someone go through the domain, range, and horizontal asympote with me? And other problems I have on the same topic

Answer 1

is your function this?
\[f(x)=3x*\left( \frac{ 1 }{ 2 } \right)^{x}\]

Answer 2

The 3 is by itself

Answer 3

\[f(x)=3*x*\left( \frac{ 1 }{ 2 } \right)^{x}\]

Answer 4

is it ok, now?

Answer 5

\[3\times(1/2)^{x}\]

Answer 6

ok! I understand

Answer 7

domain of of is the set of real numbers, namely \[\mathbb{R} \]

Answer 8

because your function is continuous everywhere in R, do you agree?

Answer 9

furthermore, since exponential function is Always positive, also your function is, so range of your function is the set:
\[[0,+\infty)\]

Answer 10

Yes I get it so far

Answer 11

Sorry I lost my connection

Answer 12

ok! in order to find horizontal asymptotes we have to calculate these limits:
\[\lim _{\pm \infty}f(x)=\lim _{\pm \infty }\frac{ 3 }{ 2^{x} }\]

Answer 13

now:
\[\lim _{x \rightarrow +\infty}f(x)=0\]
and:
\[\lim _{x \rightarrow -\infty}f(x)=+\infty\]
is it ok?

Answer 14

I don't get that part

Answer 15

is it clear for you?

Answer 16

for the first limit, keep in mind that when x-->+infinity, then 2^x--->+infinity, so:
\[\frac{ 1 }{ 2^{x} }\rightarrow 0\]

Answer 17

for second limit, keep in mind that when x-->-infinity, then 2^x-->0 through positive numbers, so:
\[\frac{ 1 }{ 2^{x} }\rightarrow + \infty\]

Answer 18

now?

Answer 19

Yes, I think I got it

Answer 20

ok!
since f(x) is continuous every where, there aren't vertical asymptotes, so we can go to find maximum and minimum points

Answer 21

first we note that:
\[f(0)=\frac{ 3 }{ 2^{0} }=3\]

Answer 22

so we can draw this:

Answer 23

now I calculate f'(x):
\[f'(x)=-\frac{ 3 }{ 2^{x} }\ln 2\]

Answer 24

now I ask you at which points we have f'(x)=0?

Answer 25

or which is the sign of f'(x)?

Answer 26

I don't really know

Answer 27

@minisweet4 since f'(x) is Always negative, in the domain of f(x), we can conclude that f(x) is always decreasing

Answer 28

furthermore there aren't any inflection points

Answer 29

I get it

Answer 30

I think that graph of your function is: