A polynomial f(x) of degree 3 for which f(1) = -1, f'(1) = 2, f''(1) = -1, and f'''(1) = -2 is given by
A) f(x) = -1 + 2(x - 1) - $\frac { 1 } { 2 }$ $( x - 1 ) ^ { 2 }$ - $\frac { 1 } { 3 }$ $( x - 1 ) ^ { 3 }$
B) f(x) = -1 + 2(x - 1) - 1 $( x - 1 ) ^ { 2 }$ - 2 $( x - 1 ) ^ { 2 }$
C) f(x) = -1 + 2x - $x ^ { 2 }$ - 2 $x ^ { 3 }$
D) f(x) = $x ^ { 3 }$ - 2 $x ^ { 2 }$ - 3x + 2
E) none of these

Question

Quizplus · Accepted Answer

Given the conditions, the polynomial can be constructed using Taylor's theorem around x=1. The coefficients of the polynomial are determined by the function's value and its derivatives at x=1. Choice A correctly applies these values: f(1)=-1, f'(1)=2, f''(1)=-1 (which gives the $\frac{-1}{2}$ factor for the quadratic term when divided by 2!), and f'''(1)=-2 (which gives the $\frac{-2}{6} = \frac{-1}{3}$ factor for the cubic term).

A Polynomial F(x) of Degree 3 for Which F(1) = 12\frac { 1 } { 2 }21​

A Polynomial F(x) of Degree 3 for Which F(1) = $\frac { 1 } { 2 }$