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 October 6th, 2012, 03:30 PM #1 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 1st derivative test For the following function, fill in the table to find where the function is increasing or decreasing. y = f(x) = x^(2) + 6x + 10 f ' (x) = 2x + 6 open interval -infinity
October 6th, 2012, 05:41 PM   #2
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Re: 1st derivative test

Quote:
 Originally Posted by mathkid y = f(x) = x^(2) + 6x + 10 f ' (x) = 2x + 6
Correct. So, we do

2x + 6 = 0

2x = -6

x = -6/2 = -3.

Note that f(x) is a parabola, so there is one (global) extremum, a minimum at x = -3, y = 1.

Good work.

 October 6th, 2012, 06:29 PM #3 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: 1st derivative test Thanks Greg
 October 6th, 2012, 06:30 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: 1st derivative test To determine the intervals on which the function is increasing/decreasing, you first find the critical number(s). For this function this is $x=-3$ This gives you the two intervals: i) $$$-\infty,-3$$$ Test value $x=-4$ $f'(-4)<0$ function decreasing on this interval. ii) $$$-3,\infty$$$ Test value $x=0$ $f'(0)>0$ function increasing on this interval. So, we can therefore conclude that the function has a relative (in this case a global) minimum at $x=-3$.
 October 6th, 2012, 07:14 PM #5 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: 1st derivative test I got it right Mark. Need help on other question finding critical value(s).
 October 6th, 2012, 07:22 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: 1st derivative test You gave different intervals...I just wanted to demonstrate the method I was taught to use.

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