Answer
$f(x)=-7+23(x+1)-41(x+2)^2+36(x+1)^3-16(x+1)^4+3(x+1)^5$
Work Step by Step
We have $f'(x)=15x^4-4x^3+6x^2+2x \implies f'(-1)=23$;
$f''(x)=60x^3-12x^2+12x+2 \implies f''(-1)=-82\\ f'''(x)=180x^2-24x+12 \implies f'''(2)=216$ and $f^{4}(-1)=-384$
Therefore, the Taylor's series at $x=-1$ is as follows:
$f(x)=f(-1)+f'(-1) (x+1)+\dfrac{f''(-1)(x+1)^2 }{2!}+\dfrac{f''(-1)(x+1)^3 }{3!}+\dfrac{f''(-1)(x+1)^4 }{4!}$
or, $f(x)=-7+23(x+1)-41(x+2)^2+36(x+1)^3-16(x+1)^4+3(x+1)^5$
