#### Answer

$f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$

#### Work Step by Step

Differentiate the given function $f(x)$ as follows:
$f'(x)=6x^2+2x+3 \implies f'(1)=11 \\ f''(x)=12x+2 \implies f''(1)=14\\ f'''(x)=12 \implies f'''(2)=12$
The Taylor's series at $x=1$ can be written as:
$f(x)=f(1)+f'(1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f'''(1)(x-1)^3 }{3!} =-2+11(x-1)+\dfrac{14}{2} (x-1)^2+\dfrac{12}{6} (x-1)^3$
So, $f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$