Método de Bisección ejercicio resuelto paso a paso PDF

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Método de Bisección��(��)= 4��2− 5�� ����= 1 ����= 1.Paso 1��( ����) ��( ����) < 0��( ����) = 4(0)2− 5( 0) = −����(����)= 4(1)2− 5(1)= ��. ����Pasó 2����=����+ ����2=1 + 1.2=��.��Pasó 3��(����) = 4(1)2− 5(1)= ��. ������(����)��(����)=(−)(0. 26)=−��. ����Iteraciones��������������(����) ��(����...

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Método de Bisección 𝑓(𝑥 ) = 4𝑥 2 − 5𝑥 Paso 1 𝑓( 𝑥 𝑖 ) 𝑓( 𝑥 𝑢 ) < 0

𝑥𝑖 = 1 𝑥𝑢 = 1.6

𝑓(𝑥𝑖) = 4(0)2 − 5 (0) = −𝟏 𝑓(𝑥𝑢 ) = 4(1.6)2 − 5(1.6) = 𝟐. 𝟐𝟒 Pasó 2

𝑥𝑟 =

𝑥𝑖 + 𝑥𝑢 1 + 1.6 = 𝟏. 𝟑 = 2 2

Pasó 3

𝑓(𝑥𝑟) = 4(1.3)2 − 5(1.3) = 𝟎. 𝟐𝟔 𝑓(𝑥𝑖 )𝑓( 𝑥𝑟) = ( −1)(0.26) = −𝟎. 𝟐𝟔 𝑥𝑖 𝑥𝑢 𝑥𝑟 Iteraciones 1 1 1.6 1.3 2 1 1.3 1.15 3 1.15 1.3 1.225 4 1.225 1.3 1.2625 5 1.225 1.2625 1.2438 6 1.2438 1.2625 1.2532 7 1.2438 1.2532 1.2485 It2 𝑥𝑢 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1 + 1.3 = 𝟏. 𝟏𝟓 𝑥𝑟 = = 2 2 𝑓 (𝑥𝑟) = 4(1.15)2 − 5(1.15) = −𝟎. 𝟒𝟔 It3 𝑥𝑖 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1.15 + 1.3 = = 𝟏. 𝟐𝟐𝟓 𝑥𝑟 = 2 2 𝑓(𝑥𝑖) = 4(1.15)2 − 5(1.15) = −𝟎. 𝟒𝟖

𝑓 (𝑥𝑟) = 4(1.225)2 − 5(1.225) = −𝟎. 𝟏𝟐𝟐𝟓

𝑓(𝑥𝑖 ) -1 -1 -0.48 -0.1225 -0.1225 -0.0308 -0.0308

𝑓 (𝑥 𝑟 ) 0.26 -0.46 -0.1225 0.0631 -0.0308 0.0160 -0.00075

𝑓(𝑥𝑖 )𝑓( 𝑥𝑟) -0.26 0.46 0.05635 -0.0077 0.0038 -0.0005 0.00002

It4 𝑥𝑖 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1.225 + 1.3 = 𝟏. 𝟐𝟔𝟐𝟓 𝑥𝑟 = = 2 2 𝑓(𝑥𝑖) = 4(1.225)2 − 5(1.225) = −𝟎. 𝟏𝟐𝟐𝟓

𝑓 (𝑥𝑟) = 4(1.2625) 2 − 5(1.2625) = 𝟎. 𝟎𝟔𝟑𝟏 It5 𝑥𝑢 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1.225 + 1.2625 = = 𝟏. 𝟐𝟒𝟑𝟖 𝑥𝑟 = 2 2 𝑓 (𝑥𝑟) = 4(1.2438) 2 − 5(1.2438) = −𝟎. 𝟎𝟑𝟎𝟖 It6 𝑥𝑖 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1.2438 + 1.2625 = 𝟏. 𝟐𝟓𝟑𝟐 𝑥𝑟 = = 2 2 𝑓 (𝑥𝑖 ) = 4(1.2438) 2 − 5(1.2438) = − 𝟎. 𝟎𝟑𝟎𝟖 𝑓 (𝑥𝑟) = 4(1.2532) 2 − 5(1.2532) = 𝟎. 𝟎𝟏𝟔𝟎 It7 𝑥𝑖 = 𝑥𝑟 𝑥𝑖 + 𝑥𝑢 1.2438 + 1.2532 = = 𝟏. 𝟐𝟒𝟖𝟓 𝑥𝑟 = 2 2 𝑓 (𝑥𝑖 ) = 4(1.2485) 2 − 5(1.2485) = − 𝟎. 𝟎𝟑𝟎𝟖 𝑓 (𝑥𝑟) = 4(1.2485) 2 − 5(1.2485) = −𝟎. 𝟎𝟎𝟕𝟓...