Đáp án:
$f(x) = (x-1)^3 - 4(x-1) - 2$
Giải thích các bước giải:
$f(x) = x^3 - 3x^2 - x + 1 \,\,\,\,\to f(1) = -2$
$+)\quad f'(x) = 3x^2 - 6x - 1\to f'(1) = -4$
$+)\quad f''(x) = 6x - 6 \,\,\qquad \to f''(1) = 0$
$+)\quad f'''(x) = 6 \,\,\qquad \qquad \to f'''(1) = 6$
$+)\quad f^{(4)}(x) = 0\qquad \qquad \to f^{(4)}(c) = 0$
Ta được:
$f(x) = f(1) + \dfrac{f'(1)}{1!}(x-1) + \dfrac{f''(1)}{2!}(x-1)^2+\dfrac{f'''(1)}{3!}(x-1)^3 +\dfrac{f^{(4)}(c)}{4!}(x-1)^4$
$\to f(x) = - 2 - 4(x-1) + 0 + (x-1)^3 + 0$
$\to f(x) = (x-1)^3 - 4(x-1) - 2$
