Question

Does the function f(x) = (x)^(1/3) have a Taylor polynomial approximation of degree 1?

1) True
2) False

Answer 1

The function f(x) =
`(x)^(^1^/^3^)` have a Taylor polynomial approximation of degree 1 is False.

Step-by-step explanation:

The function f(x) =
`(x)^(^1^/^3^)` does not have a Taylor polynomial approximation of degree 1. The Taylor polynomial of degree 1 for a function f(x) at x = a is given by P1(x) = f(a) + f’(a)(x - a). However, when we calculate the first derivative of f(x) =
`(x)^(^1^/^3^)`, we get f’(x) =
`(1/3)x^6^(^-^2^/^3^)`, and this derivative is not defined at x = 0. Therefore, the function does not have a Taylor polynomial approximation of degree 1.

Additionally, we can verify this by considering the behavior of the function near x = 0. The function f(x) =
`(x)^(^1^/^3^)`has a cusp at x = 0, which means that it does not have a well-defined tangent line at this point. As a result, it cannot be accurately approximated by a linear function in the neighborhood of x = 0. Therefore, the function f(x) =
`(x)^(^1^/^3^)` does not have a Taylor polynomial approximation of degree 1.

In conclusion, due to the undefined derivative at x = 0 and the cusp at that point, the function f(x) =
`(x)^(^1^/^3^)` does not have a Taylor polynomial approximation of degree 1.