Question

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

Answer 1

please help me out in this......

Answer 2

\[f(x) = e ^{-x}-1\]

Answer 3

is that the question?

Answer 4

yes this is the question?

Answer 5

okay, substitute and three positive values of x, say 1,2,3 and three negative values, say -1,-2,-3 in f(x)

Answer 6

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)

Answer 7

just go to google.com and type e^-1. you will get a value. try it out

Answer 8

according to google calculator e^-1 is 0.367879441. so (e^-1) -1 = -0.632120559
so at x = 1, f(x) = -0.632120559

Answer 9

similarly find out values of f(x) for x =2,3 and x = -1,-2,-3

Answer 10

use those values and make a graph such as this:

f(x)
|
|
|
|
|
|
|
|----------------------->x

Answer 11

iam new to this..thts why getting problems

Answer 12

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?

Answer 13

yes i did
but i dint got

Answer 14

okay. do you have a scientific calculator with you?

Answer 15

did you type in e^-1 exactly as I typed it? just copy e^-1 and paste it in google. and press enter.

Answer 16

yeah i did the same

Answer 17

i got now

Answer 18

okay what did you get?

Answer 19

can you help me in another question?

Answer 20

i got the same which you got

Answer 21

okay good, just do the same for other numbers and you will get your answers.

Answer 22

Yes, I can help you. But your new problem separately.

Answer 23

post your new problem separately.

Answer 24

ok
can i post it here

Answer 25

no.. post it separately. this post is too long already

Answer 26

ok

Answer 27

I dont see it zehera.

Answer 28

its not coming there

Answer 29

i think that i have poted it before thts y it not coming once again

Answer 30

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

Answer 31

is that it?

Answer 32

yes

Answer 33

so Y = 5x^-2 is equation 1
and
X = 5Y^-2 is equation 2
right?

Answer 34

yes

Answer 35

okay what is x^-2 ?

Answer 36

on google i got this but ith another equation

Answer 37

im getting positive values but for negative vl\alues iam getting confused

Answer 38

okay, ;ets make it simpler then. what is x^-2 ?

Answer 39

ok it would be greatful

Answer 40

okay. but what is x^-2?

Answer 41

its given that Y=5X^-2

Answer 42

no, tell me what x^-2 means. I dont want maths. Just tell me in plain english. What do you think x^-2 means?

Answer 43

I DONT KNOW?

Answer 44

okay. \[x ^{-2} = 1/x ^{2}\]

Answer 45

ok

Answer 46

so,
\[5x ^{-2} = 5/x ^{2}\]

Answer 47

that means,
\[Y = 5/x ^{2}\]

Answer 48

so now substitute x = 1,2,3 and -1,-2,-3 and get Y values. what is the value of Y at 0?

Answer 49

1 i think

Answer 50

how?

Answer 51

any value at log0 is 1

Answer 52

you are just asked to evaluate the first equation. Why are you taking log?

Answer 53

ok
you explain me how to do

Answer 54

if we take one value as 2 thn y=5/4

Answer 55

is it right?

Answer 56

right

Answer 57

ok like this we have to take the other two values

Answer 58

right. But the question also asks for x =0 . so what happens if you take another value as 0?

Answer 59

dont lknow

Answer 60

just put 0 in the equation. see what you get. y = 5/0. what is anything divided by 0?

Answer 61

0

Answer 62

so we have to take three values as 0,1,2

Answer 63

no, anything multiplied by 0 is 0. anything divided by zero is infinity.

Answer 64

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.

Answer 65

ok

Answer 66

so totally you have 7 values, three positive, three negative and one 0.

Answer 67

ok

Answer 68

okay. now second equation.

Answer 69

while taking negative values we get 5/-4

Answer 70

what does the problem say you should do with the second equation?

Answer 71

if x = -2, what is x ^2?

Answer 72

x^2 = x * x.
so, if x = -2, x * x = ?

Answer 73

4

Answer 74

so then for x = -2, Y = ?

Answer 75

confused

Answer 76

x = -2. Y = 5/(-2)^2

Answer 77

but you just said (-2)^2 = 4
so Y = 5/4.

Answer 78

understood?

Answer 79

but here the poer is -2 na

Answer 80

-2^-2=?

Answer 81

are you talking about equation 1 or equation 2?

Answer 82

equation2

Answer 83

okay
i was talking about equation 1

Answer 84

what does the problem say to do with equation 2?

Answer 85

same as we did in equation 1
but here we have to take negative values

Answer 86

No, pay attention. The problem says transform equation 2 to its logarthmic form

Answer 87

am i right?

Answer 88

im jst getting confused

Answer 89

you just explain me

Answer 90

i will follow it

Answer 91

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

Answer 92

the second equation is X=5y^-2

Answer 93

so you transform it to logarithms by taking log on both sides.
that is
log X = log(5y^-2)

Answer 94

that is all I can explain. You should be able to take it from there.

Answer 95

ok

Answer 96

thankyou

Answer 97

you are welcome.

Answer 98

you helped me lot