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September 28, 2025, 05:12:35 am
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Author Topic: TyErd's questions  (Read 42560 times)  Share 

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brightsky

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Re: TyErd's questions
« Reply #315 on: June 28, 2010, 12:08:42 pm »
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Quote from: TyErd on June 28, 2010, 11:57:33 am
Find the values of for which increases as increases and hence find the maximum value of .

I dont even know where to start on this one. Help?

I don't really understand the first part of the question but the second part:

Maxima and minima exists when gradient = dy/dx = 0.

Let ...(1)



To find maxima, let that equal 0.



By null factor law, either or

The former has no solutions, the second yields

Sub in into (1):

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brightsky

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Re: TyErd's questions
« Reply #316 on: June 28, 2010, 12:19:04 pm »
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Quote from: TyErd on June 28, 2010, 11:22:53 am
For the function has a minimum value at x equals?



Minimum for f(x) occurs when . So (only one within the domain) so .
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TyErd

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Re: TyErd's questions
« Reply #317 on: June 28, 2010, 12:27:47 pm »
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If im allowed to use the calculator why would I use the approximation formula in the first place?

Quote from: brightsky on June 28, 2010, 12:19:04 pm
Quote from: TyErd on June 28, 2010, 11:22:53 am
For the function has a minimum value at x equals?



Minimum for f(x) occurs when . So (only one within the domain) so .

how do you know the minimum is when ?
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TrueTears

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Re: TyErd's questions
« Reply #318 on: June 28, 2010, 12:30:20 pm »
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3+/-amplitude
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brightsky

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Re: TyErd's questions
« Reply #319 on: June 28, 2010, 12:33:06 pm »
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Quote from: TyErd on June 28, 2010, 12:27:47 pm
If im allowed to use the calculator why would I use the approximation formula in the first place?

It's just part of the methods course. Of course, you can work it out by hand and give an "exact solution" to the approximation if you want.

Quote from: TyErd on June 28, 2010, 11:22:53 am
how do you know the minimum is when ?

The minimum occurs when is the largest positive number as 3 - (largest positive number) = smallest number. If you consider the graph , the graph can only go as far up as 1. Hence this largest positive number is 1.
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TyErd

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Re: TyErd's questions
« Reply #320 on: June 28, 2010, 12:45:40 pm »
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Ohhh okay! thankyou so much
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TyErd

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Re: TyErd's questions
« Reply #321 on: June 28, 2010, 08:32:46 pm »
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Given that , find in terms of q the approximate increase in y as x increase from 0 to q.

I get the answer of 2q+1 but the answer is just 2q

I used the linear approximation formula
« Last Edit: June 28, 2010, 09:06:36 pm by TyErd »
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TyErd

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Re: TyErd's questions
« Reply #322 on: June 28, 2010, 09:10:53 pm »
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Same trouble with this one:

Fidn the approximate change in y when x changes from a to a+p, where  p is small.
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TyErd

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Re: TyErd's questions
« Reply #323 on: June 28, 2010, 09:12:35 pm »
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help please
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brightsky

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Re: TyErd's questions
« Reply #324 on: June 28, 2010, 09:27:34 pm »
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Quote from: TyErd on June 28, 2010, 09:10:53 pm
Same trouble with this one:

Fidn the approximate change in y when x changes from a to a+p, where  p is small.




As



And because x changes from a to a + p,

So
« Last Edit: June 28, 2010, 09:31:48 pm by brightsky »
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brightsky

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Re: TyErd's questions
« Reply #325 on: June 28, 2010, 09:29:54 pm »
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Quote from: TyErd on June 28, 2010, 08:32:46 pm
Given that , find in terms of q the approximate increase in y as x increase from 0 to q.

I get the answer of 2q+1 but the answer is just 2q

I used the linear approximation formula



As



As x changes from 0 to q,

So
« Last Edit: June 28, 2010, 09:31:59 pm by brightsky »
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TyErd

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Re: TyErd's questions
« Reply #326 on: June 28, 2010, 09:36:55 pm »
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Can you solve it using the formula ?
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TyErd

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Re: TyErd's questions
« Reply #327 on: June 28, 2010, 09:37:18 pm »
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Im never sure about which one to use
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brightsky

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Re: TyErd's questions
« Reply #328 on: June 28, 2010, 10:04:54 pm »
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They mean the same thing, you can use either, the change in x in the non-Leibniz notation would be h and the change in y would be f(x+h) - f(x).
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TyErd

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Re: TyErd's questions
« Reply #329 on: June 28, 2010, 10:14:10 pm »
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I tried it on the above questions but it didn't work out for me :(
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