Решебник по алгебре 11 класс Никольский Параграф 5. Применение производной Задание 67

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\[\mathbf{f}\left( \mathbf{x} \right)\mathbf{=}\mathbf{a}_{\mathbf{n}}\mathbf{x}^{\mathbf{n}}\mathbf{+}\mathbf{a}_{\mathbf{n - 1}}\mathbf{x}^{\mathbf{n - 1}}\mathbf{+}\]

\[\mathbf{+ \ldots +}\mathbf{a}_{\mathbf{1}}\mathbf{x +}\mathbf{a}_{\mathbf{0}}\mathbf{;\ \ n \geq 2}\]

\[f^{'}(x) = na_{n}x^{n - 1} +\]

\[+ (n - 1)a_{n - 1}x^{n - 2} + \ldots + a_{1};\]

\[f"(x) = n(n - 1)a_{n}x^{n - 2} +\]

\[+ (n - 1)(n - 2)a_{n - 1}x^{n - 3} +\]

\[+ \ldots + 2a_{2};\]

\[f^{'''}(x) = n(n - 1)(n - 2)a_{n}x^{n - 3} +\]

\[+ (n - 1)(n - 2)(n - 3)a_{n - 1}x^{n - 4} +\]

\[+ \ldots + 6a_{3}.\]

\[n = 2:\]

\[f^{'}(x) = 2a_{2}x^{2 - 1} +\]

\[+ (2 - 1)a_{2 - 1}x^{2 - 2} = 2a_{2}x + a_{1};\]

\[f"(x) = 2 \cdot (2 - 1)a_{2}x^{2 - 2} = 2a_{2};\]

\[f'''(x) = 0.\]

\[n = 3:\]

\[f^{'(x)} = 3a_{3}x^{3 - 1} +\]

\[+ (3 - 1)a_{(3 - 1)}x^{3 - 2} +\]

\[+ (3 - 2)a_{3 - 2}x^{3 - 3} =\]

\[= 3a_{3}x^{2} + 2a_{2}x + a_{1};\]

\[f^{''(x)} = 3 \cdot (3 - 1)a_{3}x^{3 - 2} +\]

\[+ (3 - 1)(3 - 2)a_{3 - 1}x^{3 - 3} =\]

\[= 6a_{3}x + 2a_{2};\]

\[f^{'''(x)} =\]

\[= 3 \cdot (3 - 1)(3 - 2)a_{3}x_{3 - 3} =\]

\[= 6a_{3}.\]

\[n = 4:\]

\[f^{'(x)} = 4a_{4}x^{4 - 1} +\]

\[+ (4 - 1)a_{4 - 1}x^{4 - 2} +\]

\[+ (4 - 2)a_{4 - 2}x^{4 - 3} +\]

\[+ (4 - 3)a_{4 - 3}x^{4 - 4} =\]

\[= 4a_{4}x^{3} + 3a_{3}x^{2} + 2a_{2}x + a_{1};\]

\[f^{''(x)} = 4 \cdot (4 - 1)a_{4}x^{4 - 2} +\]

\[+ (4 - 1)(4 - 2)a_{4 - 1}x^{4 - 3} +\]

\[+ (4 - 2)(4 - 3)a_{4 - 2}x^{4 - 4} =\]

\[= 12a_{4}x^{2} + 6a_{3}x + 2a_{2};\]

\[f^{'''(x)} =\]

\[= 4 \cdot (4 - 1)(4 - 2)a_{4}x^{4 - 3} +\]

\[+ (4 - 1)(4 - 2)(4 - 3)a_{4 - 1}x^{4 - 4} =\]

\[= 24a_{4}x + 6a_{3}.\]