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If `f` and `g` are inverse functions, then `g'(x) = 1/(f'(g(x)))` In the applet above, we will see a geometric justification for this formula Drag the slider through the steps and consider the questions below Explore Step 0 We are given `f` and `g` and our goal is to compute `g'(c)` JustTheorem 2 If f'(x) = g'(x) for all x in an interval (a, b) of the domain of these functions, then f g is constant or f = g c where c is a constant on (a, b Proof Let F = f − g, then F' = f' − g' = 0 on the interval (a, b), so the above theorem 1 tells that F = f − g is a constant c or f = g c Theorem 3 If F is an antiderivative of f on an interval I, then the most generalG x b N X Ђ̓e X g f B A i e X g f B X N j ̔̔ A A f B I i ̐ { n ̔̔ Ђł B ̑ Ɂy f _ A x Z Ҍ C I z p u L v u ϕi E { э܁v Љ Ă ܂ B e X g f B A ƂƂ́E E e X g f B A Ƃ́A ȕi K i ̋K ƂȂ f B X N e v Ȃǂ̑ ̂ł A F l

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(f o g)(x) = (g o f)(x) generally false for composition That is, you cannot reverse the order in composition and expect to end up with the correct result Composition is not flexible like multiplication, and is an entirely different process Do not try to multiply functions when you are supposed to be plugging them into each other You can use the Mathway widget below toIllustration ^ l C X g V @ @ @ @ @ @ l C X g V @ Illustration ^ ̑ @ @ @ @ @ @ ̑ @ Illustration ^MAZE i } E C Y j @ @ @ @ @ @ MAZE E } E C Y @ Illustration ^ C X g V X g TOP design & marketing Illustration�A X ^ E x f B A X g b p Y ł B A v X A y j R A o J R A J p ` A R ɕ z Ă ܂ B W X O O ` Q W O O g ̑ n Ȃǂɐ A ͂P T ` Q O Z ` ɂȂ ܂ B t ́A قډ~ ` 炳 ` A ȉ~ ` őS ł B�

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G(3) = 4 – (t) = 4 – t There's nothing more I can do with this, and I can't find a fully numerical value because I don't have a number to plug in for the t So my answer is g(3) = 4 – t Given that f (x) = 3x 2 2x, find f (h) Everywhere that my formula has an "x", I now plug in an "h" I start with the formula they gave me f (x) = 3x 2 2x If I want to be excruciatingly clear, IIDSONY( I X g A) SONY u h ̓ { Ή DVDR f B A B SONY ЍH ꐻ ƌ I X g A Y ̕ ł B ɏo Ĉȍ~ p Ă ܂ f B A ł A Ƃ̋߂ ̃z Z ^ CAINZ HOME i J C Y z j Ŕ c 𔭌 A ߊl ɐ ܂ B iX X g A { f B A } A ` F X g w b g A n b g A L b v A C v N ^ i T O X ̑ j x g T X y _ u c A j p b g ̑ t b g p i O u L ̓ b y ̑ o b O ` z X ^ K P X X O X R v A h

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ЊT v ̃y W B Ńz y W 삵 Ă X } g f B A ł B B ͖ ֌q f U C N T A ƂƂ q l ֏Ί 񋟂ł 悤 ɁA l X ȉ \ ͍ Ă ܂There exists x 2A such that (g f)(x) = g(f(x)) = z Therefore if we let y = f(x) 2B, then g(y) = z Thus g is surjective Problem 338 In each part of the exercise, give examples of sets A;B;C and functions f A !B and g B !C satisfying the indicated properties (a) g is not injective but g f is injective (b) f is not surjective but g f is surjective Solution The same example works for(fg)(x h) (fg)(x) = (f(x)g0(x) g(x)f0(x))h r fg(h) (19) where jr fg(h)j jhj!0 as h!0 Therefore, by the Landau de nition of di erentiability, we have shown that fgis di erentiable at every point x2Uand that its derivative is equal to f(x)g0(x)g(x)f0(x) = fDg gDf Note that this derivative is unique by Theorem 912 in Rudin 3 Let T be a linear transformation from Rn to R m Show that T

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