EVALUATE THE EXPONENTIAL FUNCTION FOR THREE POSITIVE VALUES OF X,THREE NEGATIVE VALUES OF X,AND AT X=0.SHOW YOUR WORK.USE THE RESULTING ORDERED PAIRS TO PLOT THE GRAPH.STATE THE DOMAIN AND RANGE OF THE FUNCTIONATTACH GRAPH AS LISTED AND REQUIRED.F(X)=E^X1
 anonymous
EVALUATE THE EXPONENTIAL FUNCTION FOR THREE POSITIVE VALUES OF X,THREE NEGATIVE VALUES OF X,AND AT X=0.SHOW YOUR WORK.USE THE RESULTING ORDERED PAIRS TO PLOT THE GRAPH.STATE THE DOMAIN AND RANGE OF THE FUNCTIONATTACH GRAPH AS LISTED AND REQUIRED.F(X)=E^X1
 katieb
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 anonymous
please help me out in this......
 anonymous
\[f(x) = e ^{x}1\]
 anonymous
is that the question?
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 anonymous
yes this is the question?
 anonymous
okay, substitute and three positive values of x, say 1,2,3 and three negative values, say 1,2,3 in f(x)
 anonymous
so for x =1,
f(x) = f(1) = (e^1) 1
use your calculator or google to find the value of e^1 and subtract 1 from that to get f(1)
 anonymous
just go to google.com and type e^1. you will get a value. try it out
 anonymous
according to google calculator e^1 is 0.367879441. so (e^1) 1 = 0.632120559
so at x = 1, f(x) = 0.632120559
 anonymous
similarly find out values of f(x) for x =2,3 and x = 1,2,3
 anonymous
use those values and make a graph such as this:
f(x)







>x
 anonymous
iam new to this..thts why getting problems
 anonymous
okay I understand. But now I have shown you the procedure. Just follow it. Its easy. You can do it. Did you try typing it in google?
 anonymous
yes i did
but i dint got
 anonymous
okay. do you have a scientific calculator with you?
 anonymous
did you type in e^1 exactly as I typed it? just copy e^1 and paste it in google. and press enter.
 anonymous
yeah i did the same
 anonymous
i got now
 anonymous
okay what did you get?
 anonymous
can you help me in another question?
 anonymous
i got the same which you got
 anonymous
okay good, just do the same for other numbers and you will get your answers.
 anonymous
Yes, I can help you. But your new problem separately.
 anonymous
post your new problem separately.
 anonymous
ok
can i post it here
 anonymous
no.. post it separately. this post is too long already
 anonymous
ok
 anonymous
I dont see it zehera.
 anonymous
its not coming there
 anonymous
i think that i have poted it before thts y it not coming once again
 anonymous
eveluate the exponentaial eqution forthree positive valu f x,and three negative values of x,and at x=0.transform the second expression into the equivalent logarithm equation and evalaute the logarithm equation for thre values of x that are greater than 1,three values of x that are between 0 and1,and at x=1.show your work.use the resulting ordered pairs to plot the graph of each function.Y=5x^2,X=5y^2
 anonymous
is that it?
 anonymous
yes
 anonymous
so Y = 5x^2 is equation 1
and
X = 5Y^2 is equation 2
right?
 anonymous
yes
 anonymous
okay what is x^2 ?
 anonymous
on google i got this but ith another equation
 anonymous
im getting positive values but for negative vl\alues iam getting confused
 anonymous
okay, ;ets make it simpler then. what is x^2 ?
 anonymous
ok it would be greatful
 anonymous
okay. but what is x^2?
 anonymous
its given that Y=5X^2
 anonymous
no, tell me what x^2 means. I dont want maths. Just tell me in plain english. What do you think x^2 means?
 anonymous
I DONT KNOW?
 anonymous
okay. \[x ^{2} = 1/x ^{2}\]
 anonymous
ok
 anonymous
so,
\[5x ^{2} = 5/x ^{2}\]
 anonymous
that means,
\[Y = 5/x ^{2}\]
 anonymous
so now substitute x = 1,2,3 and 1,2,3 and get Y values. what is the value of Y at 0?
 anonymous
1 i think
 anonymous
how?
 anonymous
any value at log0 is 1
 anonymous
you are just asked to evaluate the first equation. Why are you taking log?
 anonymous
ok
you explain me how to do
 anonymous
if we take one value as 2 thn y=5/4
 anonymous
is it right?
 anonymous
right
 anonymous
ok like this we have to take the other two values
 anonymous
right. But the question also asks for x =0 . so what happens if you take another value as 0?
 anonymous
dont lknow
 anonymous
just put 0 in the equation. see what you get. y = 5/0. what is anything divided by 0?
 anonymous
0
 anonymous
so we have to take three values as 0,1,2
 anonymous
no, anything multiplied by 0 is 0. anything divided by zero is infinity.
 anonymous
No, pay attention zehera. the problem asks for three positive values. so you use any three positive values. Similarly use three negative values. then use x = 0 which is neither positive nor negative.
 anonymous
ok
 anonymous
so totally you have 7 values, three positive, three negative and one 0.
 anonymous
ok
 anonymous
okay. now second equation.
 anonymous
while taking negative values we get 5/4
 anonymous
what does the problem say you should do with the second equation?
 anonymous
if x = 2, what is x ^2?
 anonymous
x^2 = x * x.
so, if x = 2, x * x = ?
 anonymous
4
 anonymous
so then for x = 2, Y = ?
 anonymous
confused
 anonymous
x = 2. Y = 5/(2)^2
 anonymous
but you just said (2)^2 = 4
so Y = 5/4.
 anonymous
understood?
 anonymous
but here the poer is 2 na
 anonymous
2^2=?
 anonymous
are you talking about equation 1 or equation 2?
 anonymous
equation2
 anonymous
okay
i was talking about equation 1
 anonymous
what does the problem say to do with equation 2?
 anonymous
same as we did in equation 1
but here we have to take negative values
 anonymous
No, pay attention. The problem says transform equation 2 to its logarthmic form
 anonymous
am i right?
 anonymous
im jst getting confused
 anonymous
you just explain me
 anonymous
i will follow it
 anonymous
There are two parts of the problem. We finished the first part using equation 1. Now the second part of the problem says transform the second equation to logarithm form
 anonymous
the second equation is X=5y^2
 anonymous
so you transform it to logarithms by taking log on both sides.
that is
log X = log(5y^2)
 anonymous
that is all I can explain. You should be able to take it from there.
 anonymous
ok
 anonymous
thankyou
 anonymous
you are welcome.
 anonymous
you helped me lot
 anonymous
thankyou very much
 anonymous
bye:)
 anonymous
http://en.wikipedia.org/wiki/Logarithms
 anonymous
you are welcome. try to understand what you are doing and you will get your answers easily. good luck!
 anonymous
thankyou
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