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finalreviewpacketsoln_pn

# finalreviewpacketsoln_pn - Calculus I Final Review...

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Unformatted text preview: Calculus I Final Review Solutions 1. In general an exponential function through ( x 1 ,y 1 ) , ( x 2 ,y 2 ) with horizontal asymptote y = a is y = a + ( y 1- a ) y 2- a y 1- a x- x 1 x 2- x 1 . (Verify this!) So, we have the equations y = 5 12 5 x- 2 2 and y = 20- 15 ( 8 15 ) x- 2 2 . Alternatively, you can find specific a nd b for the equations y = ab x and y = 20 + ab x . 2. A ( t ) = A 1 2 t k . 1 = A ( t ) A = 1 2 t 125 ln( . 1) =- t 125 ln 2 125 ln(10) ln 2 = t 3. Two examples are y = 3 + 5 sin π 6 ( x + 1) = 3 + 5 cos π 6 ( x- 2) . 4. First, each piece is a polynomial, so the only place f may not be continuous or differentiable is at x = c. So, we need to check if the function values and derivatives agree on either side of c. c 3- 2 c 2 + 3 c- 2 = 2 c- 2 0 = c 3- 2 c 2 + c = c ( c 2- 2 c + 1) = c ( c- 1) 2 So, f is continuous if c = 0 and c = 1. Now, do the derivatives agree over these c ’s? c = 0 : 3(0) 2- 4(0) + 3 = 3 6 = 2 c = 1 : 3(1) 2- 4(1) + 3 = 2 = 2 So, f is differentiable if c = 1 . 5. (a) lim x →∞ ln x 8 x 2 + 1 L’H = lim x →∞ 1 /x 16 x = lim x →∞ 1 16 x 2 = 0 . (b) lim x → + (sin x ) ln x = “0-∞ ” = “ 1 ∞ ” = “ 1 ” = ∞ (c) lim x → 1 (ln x ) ln x = lim x → 1 e ln( x ) ln(ln x ) = lim x → 1 e ln(ln x ) 1 / ln( x ) L’H = lim x → 1 e- 1 / ln( x ) · 1 /x- 1 / (ln x ) 2 · 1 /x = lim x → 1 e ln( x ) = lim x → 1 x = 1 ....
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finalreviewpacketsoln_pn - Calculus I Final Review...

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