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Use a Symbolic Differentiation Utility to Find the Fourth-Degree Taylor f(x)=1x+13f ( x ) = \frac { 1 } { \sqrt [ 3 ] { x + 1 } }

Question 52

Multiple Choice

Use a symbolic differentiation utility to find the fourth-degree Taylor polynomials (centred at zero) . f(x) =1x+13f ( x ) = \frac { 1 } { \sqrt [ 3 ] { x + 1 } }


A) S4(x) =1x3+2x2914x381+35x4243S _ { 4 } ( x ) = 1 - \frac { x } { 3 } + \frac { 2 x ^ { 2 } } { 9 } - \frac { 14 x ^ { 3 } } { 81 } + \frac { 35 x ^ { 4 } } { 243 }
B) S4(x) =1+x3+2x29+14x381+35x4243S _ { 4 } ( x ) = 1 + \frac { x } { 3 } + \frac { 2 x ^ { 2 } } { 9 } + \frac { 14 x ^ { 3 } } { 81 } + \frac { 35 x ^ { 4 } } { 243 }
C) S4(x) =1x32x2914x38135x4243S _ { 4 } ( x ) = 1 - \frac { x } { 3 } - \frac { 2 x ^ { 2 } } { 9 } - \frac { 14 x ^ { 3 } } { 81 } - \frac { 35 x ^ { 4 } } { 243 }
D) S4(x) =1+x32x29+14x38135x4243S _ { 4 } ( x ) = 1 + \frac { x } { 3 } - \frac { 2 x ^ { 2 } } { 9 } + \frac { 14 x ^ { 3 } } { 81 } - \frac { 35 x ^ { 4 } } { 243 }
E) S4(x) =1x3+2x29+14x38135x4243S _ { 4 } ( x ) = 1 - \frac { x } { 3 } + \frac { 2 x ^ { 2 } } { 9 } + \frac { 14 x ^ { 3 } } { 81 } - \frac { 35 x ^ { 4 } } { 243 }

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