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January 30, 2015

January 30, 2015

Posted by **elley** on Tuesday, November 11, 2008 at 7:35pm.

y=f(x)to find the points at which f has a

local maximum

local minimum

point of inflection

y'=(x-1)^2(x-2)(x-4)

- 12th Calculus -
**bobpursley**, Tuesday, November 11, 2008 at 7:42pmif f(x)=uvw

then

f'(x)=uvw'+ uwv'+ wvu'

- 12th Calculus -
**Damon**, Tuesday, November 11, 2008 at 7:47pmI will assume you really meant to write the derivative y' and not the function y although I would have thought the book would ask you to find that derivative.

Wherever that derivative, y', is zero, the curve is horizontal

This is twice at x = 1 and at x =2 and at x = 4

Look at the second derivative to see what happens there.

y'=(x-1)^2 (x^2-6x+8)

so

y" = (x-1)^2(2x-6)+2(x-1)(x^2-6x+8)

now what is y"at x = 1?

y"(1) = Zero, so that is an inflection point

what is y" at x = 2?

y"(2) = -2+0 = -2

That is a maximum because y" is negative

what is y"(4)?

y"(4) = 9(2)+2*3(16-24+8)

y"(4) = 18+0 = 18

so it is Minimum at x = 4

y"

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