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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 / (t-2) ; [3,5] h'(t) = -2 / (t-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) = x�, [0,1] f'(x) = 3x� = 1 => x = ?(1/3) => This my final answer? --- f(x) = (x+1) / x, [-1,2] f'(x) = -1 / x� = 1 / 2 x� = -2 => Can't square root a negative, so what's next? --------------- First Derivative Test: f(x) = (x+2)�(x-1) 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) (x-1) + (x+2)� (1) => 2x�-2x + (x+2)� ? Using a derivative calculator, I got this: 3x�+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
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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:
 [color=beige]. . [/color]

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] Tags absolute, critical, derivative, extremas, numbers, test ,

critical number for absolute value function

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