anonymous
  • 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^-X-1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
please help me out in this......
anonymous
  • anonymous
\[f(x) = e ^{-x}-1\]
anonymous
  • anonymous
is that the question?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
yes this is the question?
anonymous
  • 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
  • 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
  • anonymous
just go to google.com and type e^-1. you will get a value. try it out
anonymous
  • 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
  • anonymous
similarly find out values of f(x) for x =2,3 and x = -1,-2,-3
anonymous
  • anonymous
use those values and make a graph such as this: f(x) | | | | | | | |----------------------->x
anonymous
  • anonymous
iam new to this..thts why getting problems
anonymous
  • 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
  • anonymous
yes i did but i dint got
anonymous
  • anonymous
okay. do you have a scientific calculator with you?
anonymous
  • 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
  • anonymous
yeah i did the same
anonymous
  • anonymous
i got now
anonymous
  • anonymous
okay what did you get?
anonymous
  • anonymous
can you help me in another question?
anonymous
  • anonymous
i got the same which you got
anonymous
  • anonymous
okay good, just do the same for other numbers and you will get your answers.
anonymous
  • anonymous
Yes, I can help you. But your new problem separately.
anonymous
  • anonymous
post your new problem separately.
anonymous
  • anonymous
ok can i post it here
anonymous
  • anonymous
no.. post it separately. this post is too long already
anonymous
  • anonymous
ok
anonymous
  • anonymous
I dont see it zehera.
anonymous
  • anonymous
its not coming there
anonymous
  • anonymous
i think that i have poted it before thts y it not coming once again
anonymous
  • 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
  • anonymous
is that it?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so Y = 5x^-2 is equation 1 and X = 5Y^-2 is equation 2 right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay what is x^-2 ?
anonymous
  • anonymous
on google i got this but ith another equation
anonymous
  • anonymous
im getting positive values but for negative vl\alues iam getting confused
anonymous
  • anonymous
okay, ;ets make it simpler then. what is x^-2 ?
anonymous
  • anonymous
ok it would be greatful
anonymous
  • anonymous
okay. but what is x^-2?
anonymous
  • anonymous
its given that Y=5X^-2
anonymous
  • 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
  • anonymous
I DONT KNOW?
anonymous
  • anonymous
okay. \[x ^{-2} = 1/x ^{2}\]
anonymous
  • anonymous
ok
anonymous
  • anonymous
so, \[5x ^{-2} = 5/x ^{2}\]
anonymous
  • anonymous
that means, \[Y = 5/x ^{2}\]
anonymous
  • 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
  • anonymous
1 i think
anonymous
  • anonymous
how?
anonymous
  • anonymous
any value at log0 is 1
anonymous
  • anonymous
you are just asked to evaluate the first equation. Why are you taking log?
anonymous
  • anonymous
ok you explain me how to do
anonymous
  • anonymous
if we take one value as 2 thn y=5/4
anonymous
  • anonymous
is it right?
anonymous
  • anonymous
right
anonymous
  • anonymous
ok like this we have to take the other two values
anonymous
  • anonymous
right. But the question also asks for x =0 . so what happens if you take another value as 0?
anonymous
  • anonymous
dont lknow
anonymous
  • anonymous
just put 0 in the equation. see what you get. y = 5/0. what is anything divided by 0?
anonymous
  • anonymous
0
anonymous
  • anonymous
so we have to take three values as 0,1,2
anonymous
  • anonymous
no, anything multiplied by 0 is 0. anything divided by zero is infinity.
anonymous
  • 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
  • anonymous
ok
anonymous
  • anonymous
so totally you have 7 values, three positive, three negative and one 0.
anonymous
  • anonymous
ok
anonymous
  • anonymous
okay. now second equation.
anonymous
  • anonymous
while taking negative values we get 5/-4
anonymous
  • anonymous
what does the problem say you should do with the second equation?
anonymous
  • anonymous
if x = -2, what is x ^2?
anonymous
  • anonymous
x^2 = x * x. so, if x = -2, x * x = ?
anonymous
  • anonymous
4
anonymous
  • anonymous
so then for x = -2, Y = ?
anonymous
  • anonymous
confused
anonymous
  • anonymous
x = -2. Y = 5/(-2)^2
anonymous
  • anonymous
but you just said (-2)^2 = 4 so Y = 5/4.
anonymous
  • anonymous
understood?
anonymous
  • anonymous
but here the poer is -2 na
anonymous
  • anonymous
-2^-2=?
anonymous
  • anonymous
are you talking about equation 1 or equation 2?
anonymous
  • anonymous
equation2
anonymous
  • anonymous
okay i was talking about equation 1
anonymous
  • anonymous
what does the problem say to do with equation 2?
anonymous
  • anonymous
same as we did in equation 1 but here we have to take negative values
anonymous
  • anonymous
No, pay attention. The problem says transform equation 2 to its logarthmic form
anonymous
  • anonymous
am i right?
anonymous
  • anonymous
im jst getting confused
anonymous
  • anonymous
you just explain me
anonymous
  • anonymous
i will follow it
anonymous
  • 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
  • anonymous
the second equation is X=5y^-2
anonymous
  • anonymous
so you transform it to logarithms by taking log on both sides. that is log X = log(5y^-2)
anonymous
  • anonymous
that is all I can explain. You should be able to take it from there.
anonymous
  • anonymous
ok
anonymous
  • anonymous
thankyou
anonymous
  • anonymous
you are welcome.
anonymous
  • anonymous
you helped me lot
anonymous
  • anonymous
thankyou very much
anonymous
  • anonymous
bye:)
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Logarithms
anonymous
  • anonymous
you are welcome. try to understand what you are doing and you will get your answers easily. good luck!
anonymous
  • anonymous
thankyou

Looking for something else?

Not the answer you are looking for? Search for more explanations.