# What is the first degree Taylor polynomial of function about value c f(x)=6/x^1/3, c=8? Hello!

For a function f `( x ) ` defined and enough times differentiable on some interval `( a, b ) , ` its Taylor polynomial on the degree `n ` centered at `c in ( a , b ) ` is

`sum_( k = 0 )^n ( f^( (...

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Hello!

For a function f `( x ) ` defined and enough times differentiable on some interval `( a, b ) , ` its Taylor polynomial on the degree `n ` centered at `c in ( a , b ) ` is

`sum_( k = 0 )^n ( f^( ( k ) ) ( c ) ) / ( k ! ) ( x - c )^k .`

This polynomial approximates `f ( x ) ` to something like `( x - c ) ^ ( n + 1 ) .`

For the first degree, it becomes `f ( c ) + f ' ( c ) ( x - c ) , ` a tangent line to the graph of `f ( x ) ` at the point `x = c .`

Here, `f ( x ) = 6 / x^( 1 / 3 ) = 6 x^( - 1 / 3 ) , ` so `f ' ( x ) = - 2 x^( - 4 / 3 ) . ` The point `c ` is `8 , ` find the values:
`f ( c ) = 6 / 8^( 1 / 3 ) = 6 / 2 = 3 ,`
`f ' ( c ) = - 2 8^( - 4 / 3 ) = -2 / 2^4 = - 1 / 8 .`

This way, the Taylor polynomial is

`T ( x ) = 3 - ( x - 8 ) / 8,`

which may be rewritten as `T ( x ) = - 1 / 8 x + 4 .`

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