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Happy as happy can be.
OK, thank you. I guess I am just having trouble with the signs when raised to a fractional power, but this helps me to understand a little better. Watch this space - I'm sure that there will be more questions.
I agree with this also, though I don't know how to arrive at the final answer posted above: 1 / (-M)^1/4. Can you show me how you would solve?
OK, so if you just use an unsigned M...
As soon as I try to take the square root (twice) of
Don't I have a problem since I can't come up with a real value for a negative number? If the -1 and M cancel out the negatives, how do I arrive at an answer of
I have tried to solve it this way:
which is different than
I am trying to prove:
,by using the method where if given any negative number M, we can find a number
such that satisfies if .So we begin with
and .The text gives an answer of
.I can't quite get this to work out and feel that the signs are giving me some trouble. Can anyone show me what's happening here?
Cheers.
I stand corrected. Your teacher is correct, because with a constant like 4 you already know the sign. When there is a function with an unknown (such as x), it may be positive or negative and hence the ± sign. Sorry for the bum steer to begin with.
I am trying to solve
.The text tells me that the answer is
,Whereas I continue to arrive at
.Can someone please shed some light on this?
Cheers.
Hi Anakin,
This is because the square of a negative or positive always results in a positive.
It may help to think of it as:
Awesome - thank you for taking time to help me!
Thank you, Avon.
This is how I tried to solve it originally. Where I get stuck is on the fact that 1/6 < x < 1/2... Is it that we are choosing 1/6 from that range because it makes the denominator under
the largest possible value, and therefore makes a smaller value? Do you believe that there is a reason that the book arbitrarily chose = 1/6? When you say to bound x away from 0, is that the reason why? i.e. an x value that approaches 0 causes the function to increase without bound?Also, I used this same approach to try to solve the second problem listed in the first post. The text says that the answer is
. Does this mean an arbitrary value of 4 was used for ? If that's the case, do you have any idea why the book wouldn't state the answer as = ?Thank you in advance for any insight you can lend to this!
Cheers.
Thank you for your help, again.
Can anyone simplify the epsilon-delta proof to show how to reach the answer: δ = min{1/6, ∈/18}?
Cheers.
Thank you for helping me, TheDude.
In your solution to the first problem, you mentioned further optimizing
so that it is closer to x. I am trying to reconcile how my text book arrived at the answer .Do you know how the text arrived at this answer?
Someone please help.
I am trying to prove the following two problems using the epsilon delta method: (|f(x) - L| < ∈ if 0 < |x - a| < δ)...
1.
The text arbitrarily allows δ = 1/6 and arrives at an answer of δ = min{1/6, ∈/18}. I've done the work but can't arrive at this answer, can someone help?
2.
I don't know how to manipulate |SQRT(x)-2| into the form of |x-4| (< δ) AND arrive at the book's answer; δ = 2∈.
Please help!
Someone please help.
I am trying to prove the following two problems using the epsilon delta method: (|f(x) - L| < ∈ if 0 < |x - a| < δ)...
1.
The text arbitrarily allows δ = 1/6 and arrives at an answer of δ = min{1/6, ∈/18}. I've done the work but can't arrive at this answer, can someone help?
2.
I don't know how to manipulate |SQRT(x)-2| into the form of |x-4| (< δ) AND arrive at the book's answer; δ = 2∈.
Please help!
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