Calculus e to an exponent

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mop969

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the function f of x = e^x - e^2 / x-2

limit as x approaches 2 from both sides of e^x - e^2 / x-2

Questions to answer:
1. Domain
2. what are the values of two as you approach from both sides. The answer
must be to the 0.00001 value or smaller.
3. graph the part near x=2 and at point 2
4. tell the limit
5. discuss how you answered the problem.

Also does the e represent anything
 
Also does the e represent anything
Click to expand...

Are you in a pre-calc class?. Not to sound rude, but if you don't know what 'e' is by now you're in trouble.
 
Your not being rude this problem is for extra credit and I do not understand what the e means and how to solve the rest of the problem because of it.
 
I have looked up the value for e which is approximately 2.7.... I understand that know but how do you find the limit if you can not factor. To find the domain I took the denominator out and found out that two is not part of the domain. Tanks for any help.
 
Is this a pre calc class?. I think what you can do is, using your calculator, plug in varying values getting closer and closer to 2 from the right and from the left until you get within the accuracy mentioned.

Approaching 2 from the right would be, say, start with 2.1, then 2.01, 2.001, 2.0001, and so on

From the left, plug in 1.9, 1.99, 1.999, and so on.

This way you can see what the limit approaches. You should get the same limiting value from both directions. That means the limit exists.

The domain is everything but 2 itself. Because if x=2, then we get division by 0. You're correct.

See better now?. Tell me what you get as your limit.
 
Here is the data I got:
X Y

-3 1.46785
-2.1 1.77234
-2.01 1.80924
-2.001 1.81301
-2.0001 1.81339
-2.00001 1.81343
-2 1.81343
-1.99999 1.81343
-1.9999 1.81347
-1.999 1.81385
-1.99 1.81763
-1.9 1.85628
-1 2.34039
0 0
1 4.67077
1.99999 7.38902
1.9999 7.38869
1.999 7.38536
1.99 7.35223
1.9 7.03162
2
2.00001 7.38909
2.0001 7.38943
2.001 7.39275
2.01 7.42612
2.1 7.77114
3 12.6965
 
And if x=2 does not exist in the domain is there a hole in the graph or what happens?
 
mop969 said:
I do not understand how do I determine the limit from that data?
Click to expand...


I see two reasons why you might find that data confusing.

1) It's wrong.

2) You don't seem to understand for what the exercise is asking.

The limit concept describes what is happening to the value of f(x) as the value of x gets very close to 2.

Why did you post all those values for f(x) from x = -3 to x = 3? It leads me to wonder whether or not you understand the concept of a limit.

In this exercise, we only care about values of x that are very close to 2.

(Galactus gave you these values, explicitly.)

Go back to the drawing board, and evaluate f(x) for the following values only.

1.9
1.99
1.999
1.9999
2.0001
2.001
2.01
2.1

I'm not sure what technology you're using, but be sure to let the machine know the order of operations by using parentheses, if you're entering the entire function definition or expression all at once. The numbers in your first attempt are screwy.

Then, looking at the correct values for f(x), think about the general location of the points that you would plot, if you were to plot these eight (x,f(x)) coordinates.

 
I posted that data because it is the data I used to graph the function. So, what happens since x=2 is not part of the domain is there a whole when I draw the graph or what happens? Also, I used a calculator which had e^x as a button for the e value which I believe to be the correct button if not let me know thanks.
 
How can you tell if a hole is there or a vertical asymptote? What is the limit since this function I believe has a hole?
 
mop969 said:
I posted that data because it is the data I used to graph the function.

I think you're missing my point. The instruction that you posted asks for the graph near 2. If you would like to graph over the domain [-3, 3], instead, then that extra work is fine. However, you don't need to share all that data with us. It's not relevant.

So, what happens since x=2 is not part of the domain is there a whole when I draw the graph or what happens?

You tell me. Look at your graph! What is happening to the value of f(x) as x approaches 2 from the left? What is happening to the value of f(x) as x approaches 2 from the right?
Click to expand...


Nobody here can tell you if you're still not using your calculator properly unless you show your work.

Here are the values that I get. Do they match yours?

f(1.9) = 80.6
f(1.99) = 746.2
f(1.999) = 7396.4
f(1.9999) = 73897.9
f(2.0001) = -73883.2
f(2.001) = -7381.7
f(2.01) = -731.4
f(2.1) = -65.7

 
no these are not the value I got how did you obtain these value. did you plug 1.9 in for the x in the equation then solve for f of x.

e[sup:ye5nprks]x[/sup:ye5nprks] - e[sup:ye5nprks]2[/sup:ye5nprks]/ x-2
 
mop969 said:
… [How] did you obtain these [values?] [Did] you plug 1.9 in for the x in the equation [and] then solve for f of x[?]
Click to expand...


Yes, that's what I did, except that I would use the word "evaluate" instead of "solve" to describe it.

First, some comments about terminology and notation.

If we substitute Real numbers for variables within some mathematical expression, and then do the arithmetic to find the value of that expression, then we are evaluating the expression.

If we try to find the value of some variable in an equation, then we are solving for that variable.

Also, we write "f of x" as f(x). This is called "function notation".

Second, here's my work for evaluating f(1.9):

e^1.9 = 6.6859

e^2 = 7.3891

1.9 - 2 = -0.1

f(1.9) = e^(1.9) - e^2/(1.9 - 2)

= 6.6859 - 7.3891/(-0.1)

= 6.6859 + 73.8906

= 80.5765

= 80.6

 
I am not sure if this will make a difference but the whole function is over x-2.
 
If I may interject. Is this your function?:

\(\displaystyle \frac{e^{x}-e^{2}}{x-2}\)

Please use proper grouping symbols. The way you have it written is \(\displaystyle e^{x}-\frac{e^{2}}{x-2}\)

If it is the former, then \(\displaystyle \frac{e^{1.999}-e^{2}}{1.999-2}=7.38536280...\)

\(\displaystyle \frac{e^{1.99999999}-e^{2}}{1.9999999-2}=7.289056...\)

See?. The closer you get to 2 from the left, the closer the limit gets to 7.289056...

Which is \(\displaystyle e^{2}\).

You should get the same approaching from the right. That is, if you use 2.01, 2.00000000001, and so on.
 
mop969 said:
I am not sure if this will make a difference but the whole function is over x-2.
Click to expand...


You're not using these words as if you understand what they mean. I'll guess that you're trying to tell me that the difference of exponential terms is over x - 2.

The difference of exponential terms is not "the whole function".

Of course changing the definiton of function f makes a difference! It makes a HUGE difference. (Maybe, you're in the wrong class.)

f(x) = (e^x - e^2)/(x - 2)

Graph f(x) near x = 2 by drawing a smooth curve and putting an open circle (i.e., the "hole") at the point where x = 2.

 
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