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F g A → Ris defined by (f g)(x) = f(x) g(x) Proposition 212 Suppose that f,g A → R and f ≤ g If g is bounded from above then sup A f ≤ sup A g, and if f is bounded from below, then inf A f ≤ inf A g Proof If f ≤ g and g is bounded from above, then for every x ∈ A f(x) ≤ g(x) ≤ sup A g Thus, f is bounded fromA Z h j h \ Z g k l v a Z j h a j Z o m g d Z f b a x ^ ` _ l h f A Z j _ } k l j h \ Z g b c i Z c h \ b c d Z i l Z eMaxwell's equations in integral form GG ∫Eda ⋅ =4πQ (Gauss's law Q is charge enclosed by surface S) S GG 1 ∂φB G ∫Eds ⋅ =−=emf (Faraday's law φB is B flux through surface bounded by C) C ct∂ s G ⋅ G = 4π I 1 ∂φE ∫ (Ampere's law I is current enclosed by contour C;



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CLAY DARKSTALKER DRAGONET ICEWINGS MUDWINGS NIGHTWINGSOtherwise, if f gand gis bounded from above, then f(x) g(x) sup A g for every x2A Thus, fis bounded from above by sup A g, so sup A f sup A g Similarly, f g implies that sup A( f) sup A ( g), so inf f inf g Note that f gdoes not imply that sup Af inf g;Note that (f B g)(x) ≠ (g B f)(x) This means that, unlike multiplication or addition, composition of functions is not a commutative operation The following example will demonstrate how to evaluate a composition for a given value Example 6 Find (f B g)(3) and (g B f)(3) if f ( x ) = x 2 and g ( x ) = 4 – x2 Solution Step 1 Find (f



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