# 9thsol - ln x ⇒ f x = ln x x 1 x = ln x 1 Critical numbers f x = 0 ⇒ ln x 1 = 0 ⇒ ln x =-1 ⇒ x = e-1 is a C.N The domain of ln is(0-∞ so

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Calculus 119 9th Week 1 1. Use the subsutitution t = 3 2 - x . Since t = 3 2 - x + when x 2 - , lim x 2 - e 3 2 - x = lim t + e t = 2. y 0 = d dx ( e k tan x ) = e k tan x d dx ( k tan x ) = e k tan x k (sec 2 x ) d ( x ) dx = ke k tan x sec 2 x 1 2 x = k 2 x e k tan x sec 2 x 3. By using subsutitution u = x du = 1 2 x dx Z e x x dx = Z e u (2 du ) = 2 e u + c = 2 e x + c 4. ( f - 1 ) 0 (4) = 1 f 0 ( f - 1 (4)) Let’s ﬁnd f - 1 (4) f - 1 (4) = x f ( x ) = 4 3 + x + e x = 4 x = 0 and f 0 ( x ) = 1 + e x f 0 (0) = 2 so, ( f - 1 ) 0 (4) = 1 f 0 (0) = 1 2 5. 2 ln x = ln 2+ln(3 x - 4) ln x 2 = ln 2+ln(3 x - 4) ln x 2 - ln(3 x - 4) = ln 2 ln x 2 3 x - 4 = ln 2 x 2 3 x - 4 = 2 x 2 - 6 x +8 = 0 x = 4 or x = 2 1 Please email all corrections and suggestions to these solutions to [email protected] All solutions are available on the web at the url http://www.metu.edu.tr/ htor. 1

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6. lim x →∞ [ln(2 + x ) - ln(1 + x )] = lim x →∞ ln( 2 + x 1 + x ) Note that as x → ∞ , 2+ x 1+ x 1 lim x →∞ [ln(2 + x ) - ln(1 + x )] = ln 1 = 0 7. x y = y x ln x y = ln y x y ln x = x ln y Using implicite derivative, y 0 ln x + y 1 x = ln y + x y 0 y y 0 = ln y - y x ln x - x y 8. f ( x ) = x
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Unformatted text preview: ln x ⇒ f ( x ) = ln x + x 1 x = ln x + 1 Critical numbers; f ( x ) = 0 ⇒ ln x + 1 = 0 ⇒ ln x =-1 ⇒ x = e-1 is a C.N The domain of ln is (0 ,-∞ ), so D ( f ) = (0 ,-∞ ) 1 e ∞ f ( x )-+ f ( x ) & % f ( 1 e ) = 1 e ln 1 e =-1 e . Thus, ( 1 e ,f ( 1 e )) = ( 1 e ,-1 e ) absolute minimum point. 9. By using subsititution u = ln x ⇒ du = 1 x dx Z 6 e dx x ln x = Z ln 6 ln e du u = (ln u ) | ln 6 1 = ln(ln 6)-ln 1 = ln(ln 6) 2 10. The graph of y = 1 √ x +1 is the following V olume = Z 1 π ( 1 √ x + 1 dx = Z 1 π 1 x + 1 dx = π ln( x + 1) | 1 = π ln 2-ln 1 = π ln 2 3...
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## This note was uploaded on 03/31/2010 for the course MATHEMATIC 119 taught by Professor Muhiddinuğuz during the Fall '08 term at Middle East Technical University.

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9thsol - ln x ⇒ f x = ln x x 1 x = ln x 1 Critical numbers f x = 0 ⇒ ln x 1 = 0 ⇒ ln x =-1 ⇒ x = e-1 is a C.N The domain of ln is(0-∞ so

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