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Yes one-to-one, the graph passes the horizontal test. Not one-to-one, the graph fails the horizontal test. Not one-to-one since for example the horizontal line y intersects the graph twice.

Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Step 1: y Step 2: y. Section 7. Let x" x be two numbers in the domain of an increasing function f. In either case, f x" f x and f is one-to-one.

Similar arguments hold if f is decreasing. The function g x is also one-to-one. Therefore g x is one-to-one as well. The function h x is also one-to-one. The reasoning: f x is one-to-one means that if x" x then f x" f x , so f x"" f x" , and therefore h x" h x. The composite is one-to-one also. The reasoning: If x" x then g x" g x because g is one-to-one. Since g x" g x , we also have f g x" f g x because f is one-to-one. Thus, f g is one-to-one because x" x f g x" f g x. Yes, g must be one-to-one.

If g were not one-to-one, there would exist numbers x" x in the domain of g with g x" g x. For these numbers we would also have f g x" f g x , contradicting the assumption that f g is one-to-one.

The first integral is the area between f x and the x-axis over a x b. The second integral is the area between f x and the y-axis for f a y f b. The sum of the integrals is the area of the larger rectangle with corners at 0 0 , b 0 , b f b and 0 f b minus the area of the smaller rectangle with vertices at 0 0 , a 0 , a f a and 0 f a. Therefore, W t S t for all t [a b].

See section 2. Therefore, the absolute minimum occurs at x x 2 with f " f 2 cos ln 2 , and the absolute maximum occurs at x 1 with f 1 1. This error is! From the graph an upper bound for the error is 0. Note from the graph that 0. This is consistent with the estimate given in part b above. For all positive values of x,.

The function f x 2esin x2 has a maximum whenever sin. The maximum is 2e 5. Since f x 0 for all x, f 3 0 is also an absolute minimum. Note that y ln x and ey x are the same curve; '1 ln x dx area under the curve between 1 and a; a. Let O original sound level 10 log10 aI 10" b db from Equation 6 in the text. This is no accident, because xln 2 eln 2 ln x aeln 2 bln x 2ln x.

Then, x1 2, x2 2. Many other methods may be used. The tangent line also passes throughap ln pb ln p "p p " p e, and the tangent line equation is y "e x. Therefore, y ln x lies below the graph of. Therefore, e1 is bigger. Therefore y! Thus, y y! Now 10,eln t 0. Learn more about Scribd Membership Home. Much more than documents.

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Uploaded by yo Date uploaded Jun 08, Did you find this document useful? Is this content inappropriate? Report this Document. Flag for Inappropriate Content. Download Now. Related titles. Carousel Previous Carousel Next. Jump to Page. Search inside document. The graph is symmetric about y x. Hence faxb is either always increasing or always decreasing. Let u sin t du cos t dt; t ' cot t dt ' Thus, lim yw! Note that y ln x and ey x are the same curve; '1 ln x dx area under the curve between 1 and a; a '0ln a ey dy area to the left of the curve between 0 and ln a.

The sum of these areas is equal to the area of the rectangle The tangent line passes through a! Thus after another 14 hrs, A eln Alexander Jose Auqui Tovar.

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## Matematica1 Com Razonamiento Logico 150 Problemas Resueltos

Yes one-to-one, the graph passes the horizontal test. Not one-to-one, the graph fails the horizontal test. Not one-to-one since for example the horizontal line y intersects the graph twice. Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Step 1: y Step 2: y.

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## AnĂ¡lisis Vectorial

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