6_1_6_2_Presentation_0

6_1_6_2_Presentation_0 - SECTION 6.1 AND 6.2 COMPOSITE...

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S ECTION 6.1 AND 6.2: C OMPOSITE O F UNCTIONS ; O NE TO O NE F UNCTIONS AND I NVERSE NCTIONS F UNCTIONS Math&141 Pre-Calculus I
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Suppose that an oil tanker is leaking oil and we want to be able to determine the area of the circular oil patch around the ship. It is determined that the oil is leaking from the tanker in such a way that the radius of the circular oil patch around the ship is increasing at a rate of 3 feet per minute.
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C OMPOSITE F UNCTION | Given two functions f and g, the composite function, denoted by f º g (f composed with g) is defined by: )(x) = f(g(x)) | (f º g)(x) f(g(x)) | So on the previous slide: | r(t) = 3t (radius of the spill is 3 feet per minute x time in minutes) | A(r) = π r 2 (Area is π x radius squared) (t) = (A (t) = A(r(t)) = A(3t) = ( t 2 = t 2 | A(t) (A º r)(t) A(r(t)) A(3t) π( 3t ) π 9t | A(t) = 9 π t 2 | We have skipped the “middle man” and are going straight from time to area without calculating the radius.
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C OMPOSITE F UNCTION V ISUAL | f(x) = 2x 2 + 3 g(x) = 4x 3 + 1 | (f º g)(1) | (f º g)(1) = (f(g(1)) | g(1) = 4(1) 3 + 1 = 5 | f(5) = 2(5) 2 + 3 = 2·25 + 3 = 53 | (g º f)(1) (1) = (g(f(1)) | (g º f)(1) = (g(f(1)) | f(1) = 2(1) 2 + 3 = 5 (5) = 4(5) 3 1 = 4·125 + 1 = 501 | g(5) 4(5) + 1 4 125 + 1 501
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C OMPOSITE F UNCTIONS | f(x) = 2x 2 + 3 g(x) = 4x 3 + 1 | (f º f)(-2) | (f º f)(-2) = (f(f(-2)) | f(-2) = 2(-2) 2 + 3 = 2·4 + 3 = 11 | f(11) = 2(11) 2 + 3 = 2·121 + 3 = 245 | (g º g)(-1) | (g º g)(-1) = (g(g(-1)) ( ) = 4( ) 1 = + 1 = | g(-1) = 4(-1) 3 + 1 = -4 + 1 = -3 | g(-3) = 4(-3) 3 + 1 = 4(-27) + 1 = -107 )(x) | (f º g)(x) | (f º g)(x) = (f(g(x)) | (f(g(x)) = f(4x 3 + 1) = 2(4x 3 + 1) 2 + 3 | = 2(4x 3 + 1)(4x 3 + 1) + 3 = 2(16x 6 + 8x 3 + 1) + 3 | = 32x 6 + 16x 3 + 2 + 3 = 32x 6 + 16x 3 + 5
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C OMPOSITE F UNCTIONS | f(x) = 2x 2 + 3
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6_1_6_2_Presentation_0 - SECTION 6.1 AND 6.2 COMPOSITE...

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