Help with functions
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From: Fred Nurk on 6 Jul 2010 00:36 Am I right in not being able to do the multiple-choice question at http:// sites.google.com/site/xtheunknown0/maths/differentiation? With regards to the extended response question at http://sites.google.com/ site/xtheunknown0/maths/functions, can you please show me why there is a horizontal asymptote at y = -1 / 4? TIA
From: Arturo Magidin on 6 Jul 2010 01:00 On Jul 5, 11:36 pm, Fred Nurk <albert.xtheunkno...(a)gmail.com> wrote: > Am I right in not being able to do the multiple-choice question at http:// > sites.google.com/site/xtheunknown0/maths/differentiation? HINT the FIRST: 1/sqrt(2x-3) = (2x-3)^{-1/2} HINT the SECOND: Chain Rule. HINT the THIRD: Power Rule. > With regards to the extended response question athttp://sites.google.com/ > site/xtheunknown0/maths/functions, can you please show me why there is a > horizontal asymptote at y = -1 / 4? Which function? You have f, you have g, and you are asked about f^{-1}, g^{-1}, g o f, and (g o f)^{-1}. Which graph is it you think has a horizontal asymptote at y = -1/4? -- Arturo Magidin
From: Fred Nurk on 6 Jul 2010 01:52 Arturo Magidin wrote: > <snip> > HINT the SECOND: Chain Rule. > <snip> Yes - I don't know why I didn't do it right last night... > <snip> > Which graph is it you think has a horizontal asymptote at y = -1/4? Sorry for being unclear; g o f; back to my original question...
From: Arturo Magidin on 6 Jul 2010 02:02 On Jul 6, 12:52 am, Fred Nurk <albert.xtheunkno...(a)gmail.com> wrote: > Arturo Magidin wrote: > > <snip> > > HINT the SECOND: Chain Rule. > > <snip> > > Yes - I don't know why I didn't do it right last night... > > > <snip> > > Which graph is it you think has a horizontal asymptote at y = -1/4? > > Sorry for being unclear; g o f; back to my original question... Do you know how to check for horizontal asymptotes? Hint: it has to do with limits. -- Arturo Magidin
From: Fred Nurk on 6 Jul 2010 02:19
Arturo Magidin wrote: > <snip> > Do you know how to check for horizontal asymptotes? Hint: it has to do > with limits. Aha! I couldn't think of this last night: As x -> infinity, 1 / e ^ x approaches zero. 4 * e ^ -x approaches zero so g o f(x) approaches 1 / -4 = -1 / 4 for large values of x. Thank you so much - you have made my day! |