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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)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(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)² (1) => 2x²-2x + (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
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Re: Critical Numbers / Absolute Extremas / First Derivative

Hello, RMG46!

Here is some help . . .

Quote:
 $\text{Find the critical numbers of: }\;f(x) \:=\: 2\,\!\sec x \,+\, \tan x,\;\;0\,<\,x\,=<\,2\pi=$

$\text{Note that: }\:x \:\ne\: \frac{\pi}{2},\,\frac{3\pi}{2}$

$\text{Set }f'(x)\text{ equal to zero and solve.}$

$f'(x) \;=\;2\,\!\sec x\tan x\,+\,\sec^2x \:=\:0$

[color=beige]. . . . . . . [/color]$\sec x\,\!(2\,\!\tan x\,+\,\sec x) \:=\:0$

And we have two equations to solve.

[color=beige]. . [/color]$\sec x \:=\:0 \;\text{ . . . impossible!}$

[color=beige]. . [/color]$2\tan x\,+\,\sec x \:=\:0 \;\;\;\Rightarrow\;\;\;\frac{2\sin x}{\cos x}\,+\,\frac{1}{\cos x} \:=\:0 \;\;\;\Rightarrow\;\;\;2\,\!\sin x \,+\,1\:=\:0 \;\text{ (since }\cos x \,\ne\,0)$

[color=beige]. . . . [/color]$\sin x \:=\:-\frac{1}{2} \;\;\;\Rightarrow\;\;\;x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6}$

Quote:
 $\text{Locate the absolute extremes of the function on the closed interval:}$ [color=beige]. . [/color]$h(t) \:=\:\frac{t}{t\,-\,2}\;\;[3,\,5]$

$\text{Your derivative is correct: }\:h'(t) \:=\: -\frac{2}{(t\,-\,2)^2}$

You are right . . . The graph has a vertical asymptote at $t= 2$
[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]$\begin{Bmatrix}f(3) \,=\,3 &\text{ is the absolute maximum} \\ \\ \\ f(5) \,=\,\frac{5}{3} &\text{ is the absolute minimum} \end{Bmatrix}=$

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