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March 6, 2015

March 6, 2015

Posted by **Bill** on Wednesday, October 24, 2012 at 1:09pm.

For the second derivative I got f(x)=6x-16 and then got x=16/6 when set to zero.

- Calculus -
**Damon**, Wednesday, October 24, 2012 at 1:35pm3 x^2 -16 x + 5 = 0

first derivative

at x = {16 +/-sqrt(256-60)]/6

or [16+/-14]/6 = 5 or 0.3333

second derivative

6 x - 16

is that + or - at the two points where the slope is zero?

at x = .333

2 - 16 is negative so maximum there

at x = 5

6x - 16 = 30-16 is positive so minimum there

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