Author Topic: diff q no.2 (Read 945 times) Tweet Share

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: ?

TrueTears · « **Reply #1 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]

d0minicz · « **Reply #2 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.

naved_s9994 · « **Reply #3 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...?

TrueTears · « **Reply #4 on:** July 05, 2009, 06:36:28 pm »

you'd be finding y coordinates

naved_s9994 · « **Reply #5 on:** July 06, 2009, 08:51:07 am »

ahh lol soz,

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

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Author Topic: diff q no.2 (Read 945 times) Tweet Share

naved_s9994

diff q no.2

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