The Taylor polynomial of order 2 for2( ) 1f x x x   at1a  is (9) .(21 x x  在1a  的 2 次泰勒多項式)

The question seems to be asking for the second order Taylor polynomial for the function f(x) = 2x^2 + x + 1 at the point a = 1.

The Taylor series of a function about a point can be found using the formula:

f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + ...

Here, f(x) = 2x^2 + x + 1, so we can compute the derivatives:

f'(x) = 4x + 1, f''(x) = 4.

We evaluate these at a = 1:

f(1) = 21^2 + 1 + 1 = 4, f'(1) = 41 + 1 = 5, f''(1) = 4.

So the second order Taylor polynomial for f about a = 1 is:

P_2(x) = f(1) + f'(1)(x-1) + f''(1)(x-1)^2/2! = 4 + 5(x-1) + 4(x-1)^2/2 = 4 + 5x - 5 + 2x^2 - 4x + 2 = 2x^2 + x + 1.

So, the Taylor polynomial of order 2 for f(x) = 2x^2 + x + 1 at a = 1 is indeed 2x^2 + x + 1.