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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: naved_s9994 on July 05, 2009, 06:12:26 pm

Title: diff q no.2
Post by: naved_s9994 on July 05, 2009, 06:12:26 pm: The maximum and minimum values respectively of y= x^3 - 5x^2 + 7x + 3 , 0 < x < 2.5

are: ? :'(
Title: Re: diff q no.2
Post by: TrueTears on July 05, 2009, 06:15:47 pm: differentiate

set to 0

solve for the given domain.

[since this is methods you can draw a gradient chart to see which is max which is min or beyond the methods course just use the 2nd derivative to decide whether a point is max/min]
Title: Re: diff q no.2
Post by: d0minicz on July 05, 2009, 06:16:56 pm: $\frac {dy}{dx} = 3x^2 - 10x + 7$
Let $\frac {dy}{dx} = 0$

= $3x^2 -10x + 7 =0$
= $(x-1)(3x-7)$
$x = 1 , \frac {7}{3}$

and use gradient chart to find which is Min and which is Max.
Title: Re: diff q no.2
Post by: naved_s9994 on July 05, 2009, 06:35:49 pm: yea same i got 1 and 7/3....but answers dont say so... im lostt

Q. What would i be doing if i subbed 1 and 7/3 into main equation...?
Title: Re: diff q no.2
Post by: TrueTears on July 05, 2009, 06:36:28 pm: you'd be finding y coordinates
Title: Re: diff q no.2
Post by: naved_s9994 on July 06, 2009, 08:51:07 am: ahh lol soz,

DAmn, i need sleep! lol
ill think on this again tomm