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October 23rd, 2011, 04:40 PM   #1
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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?

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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?

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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?
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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]

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