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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: VCE123456789 on November 04, 2010, 01:23:48 pm

Title: SAMPLE VCAA HELP URGENT
Post by: VCE123456789 on November 04, 2010, 01:23:48 pm: FOr question 8 i dont no how to determine its nature could someone please explain in detail .
Cheers
Title: Re: SAMPLE VCAA HELP URGENT
Post by: VCE123456789 on November 04, 2010, 01:40:29 pm: ANYONE
Title: Re: SAMPLE VCAA HELP URGENT
Post by: schnappy on November 04, 2010, 01:47:11 pm: YOU COULD EITHER USE THE SECOND DERIVATIVE TEST WHICH IS NOT ON THE COURSE OR USE THAT STUPID SIGN TABLE WHERE YOU LOOK AT THE GRADIENT ON EITHER SIDE OF THE S.P.
Title: Re: SAMPLE VCAA HELP URGENT
Post by: Linkage1992 on November 04, 2010, 01:59:26 pm: Try subbing in n-1 and n + 1 into the equation to see the gradient you get on each side.
Title: Re: SAMPLE VCAA HELP URGENT
Post by: Blakhitman on November 04, 2010, 03:58:49 pm: ${h}'(x)=\frac{x^{n-1}(n-x)}{e^x}$

Think about it if 0<x<n then $(n-x)>0$ so ${h}'(x)>0$ and if x>n then $(n-x)<0$ so ${h}'(x)<0$

using sign table you get what andrewloppol posted above.
Title: Re: SAMPLE VCAA HELP URGENT
Post by: stonecold on November 04, 2010, 04:01:33 pm: Quote from: Blakhitman on November 04, 2010, 03:58:49 pm
${h}'(x)=\frac{x^{n-1}(n-x)}{e^x}$

Think about it if 0<x<n then $(n-x)>0$ so ${h}'(x)>0$ and if x>n then $(n-x)<0$ so ${h}'(x)<0$

using sign table you get what andrewloppol posted above.

sweet. explained like a pro. makes perfect sense :D
Title: Re: SAMPLE VCAA HELP URGENT
Post by: Whatlol on November 04, 2010, 04:05:39 pm: Quote from: andrewloppol on November 04, 2010, 03:24:25 pm
make it do this
_
/ \

pr0