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October 23rd, 2011, 04:40 PM  #1 
Member Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0  Critical Numbers / Absolute Extremas / First Derivative Test
So I'm having trouble with trigonometry and finding critical numbers. Find any critical numbers of the function: f(x) = 2sec(x) + tan(x) ; 0 < x < 2?  Locate the absolute extreme of the function on the closed interval: h(t) = t / (t2) ; [3,5] h'(t) = 2 / (t2)² I think I differentiated this correctly, but wouldn't this mean that t cannot equal 2 because it's undefined? So what next next?  Mean Value Theorem: (f(b)  f(a)) / (b  a) f(x) = x³, [0,1] f'(x) = 3x² = 1 => x = ?(1/3) => This my final answer?  f(x) = (x+1) / x, [1,2] f'(x) = 1 / x² = 1 / 2 x² = 2 => Can't square root a negative, so what's next?  First Derivative Test: f(x) = (x+2)²(x1) I'm not entire sure how to find the first derivative on this. I assume you use the exponential rule and get: 2x(1), but you also need to use multiplicity rule: der.1st(2nd) + der.2nd(1st). So does that mean: 2x(1) (x1) + (x+2)² (1) => 2x²2x + (x+2)² ? Using a derivative calculator, I got this: 3x²+6x Increasing on (?,2)U(0,?) Decreasing on (2,0) (2,0) = Relative Max. (0,4) = Relative Min. So basically I need the breakdown on differentiating the original equation, and I'm assuming the rest is correct.  f(x) = x^(2/3)  4 f'(x) = 2 / x^(1/3) (X cannot equal 0) Increasing = (0,?), Decreasing on (?,0) There's a 'sharp edge' on (0,4) so that means it's not a Relative Min. correct? 
October 24th, 2011, 10:00 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Critical Numbers / Absolute Extremas / First Derivative Hello, RMG46! Here is some help . . . Quote: [color=beige]. . . . . . . [/color] And we have two equations to solve. [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . . . [/color] Quote:
You are right . . . The graph has a vertical asymptote at [color=beige]. . [/color]But we can ignore it . . . It is not in the designated interval. We see that the derivative is never zero . . . There are no horizontal tangents. Note that the derivative is always negative . . . The function is always decreasing. Therefore:[color=beige] .[/color]  

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