Answer
$f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$
Work Step by Step
We have $f(x)=2x^3+x^2+3x-8 \implies f(1)=-2$; $f'(x)=6x^2+2x+3 \implies f'(1)=11$;
$f''(x)=12x+2 \implies f''(1)=14\\ f'''(x)=12 \implies f'''(2)=12$
Therefore, the Taylor's series at $x=1$ is as follows:
$f(x)=f(1)+f'(1) (x-1)+\dfrac{f''(1)(x-1)^2 }{2!}+\dfrac{f'''(1)(x-1)^3 }{3!}$
or, $=-2+11(x-1)+\dfrac{(14) (x-1)^2 }{2 }+\dfrac{(12)(x-1)^3 }{6}$
or, $f(x)=-2+11(x-1)+7(x-1)^2+2(x-1)^3$
