Find the (a) approximate true error, (b) relative approximate error, (c) absolute relative approximate error of...

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Find the (a) approximate true error, (b) relative approximate error, (c) absolute relative approximate error of the following: 1. The derivative of ?(?)=3+2+1 at? = 0.5 when h = 0.005 and h??????? = 0.05 ??????? In the Practice 1, the approximate value of ?(?) - =3+2+1 at? = 0.5 and h = 0.05 is 2.8275 - ?(?)=> ?(?+ h) 0 h For? 0.5 when h = 0.005 ?(?)= ?(?) 2(0.5 +0.005) - (0.5) 0.005 ? (0.505) (0.5) 0.005 = Since ?(?)=3+2+1 ?(?) [(0.505 )32(0.505) + 1] [(0.5)3 + 2(0.5) + 1] - 0.005 = 2.13878 2.125 - 0.005 2(2) = 0.01387 ?(?) = 0.005 ?(?)=2.756 = - = 2.756 - 2.8275 = -0.0715 - 7.15% Relative approximate error = x 100% -0.0715 x 100% 2.756 2, 0.02594339 x 100 % 2,2.594339 % With the absolute relative approximate error of |= -2.594339 | = 2.594339 % Find the (a) approximate true error, (b) relative approximate error, (c) absolute relative approximate error of the following: 1. The derivative of ?(?)=3+2+1 at? = 0.5 when h = 0.005 and h??????? = 0.05 ??????? In the Practice 1, the approximate value of ?(?) - =3+2+1 at? = 0.5 and h = 0.05 is 2.8275 - ?(?)=> ?(?+ h) 0 h For? 0.5 when h = 0.005 ?(?)= ?(?) 2(0.5 +0.005) - (0.5) 0.005 ? (0.505) (0.5) 0.005 = Since ?(?)=3+2+1 ?(?) [(0.505 )32(0.505) + 1] [(0.5)3 + 2(0.5) + 1] - 0.005 = 2.13878 2.125 - 0.005 2(2) = 0.01387 ?(?) = 0.005 ?(?)=2.756 = - = 2.756 - 2.8275 = -0.0715 - 7.15% Relative approximate error = x 100% -0.0715 x 100% 2.756 2, 0.02594339 x 100 % 2,2.594339 % With the absolute relative approximate error of |= -2.594339 | = 2.594339 %

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