Ở bài 19 cho:
\(\left\{ \matrix{ b = - {1 \over 2} \hfill \cr c = 0 \hfill \cr d = - {3 \over 2} \hfill \cr} \right.\)
suy ra: \(f(x) = {x^3} - {1 \over {2}}{x^2} - {3 \over 2}(C)\)
a) Ta có:
\(\eqalign{
& {x_0} = - 1 \Rightarrow {y_0}={( - 1)^3} - {1 \over 2}{( - 1)^2} - {3 \over 2} = - 3 \cr
& f'(x) = 3{x^2} - x \Rightarrow f'(-1) = 3.(-1)^2 -(- 1) = 4 \cr} \)
Vậy phương trình tiếp tuyến của (C) tại \(x_0= -1\) là:
\(y + 3 = 4(x + 1) ⇔ y = 4x + 1\)
b) Ta có:
\(\eqalign{
& f'(\sin x) = 0 \cr
& \Leftrightarrow 3.{\sin ^2}x - \sin x = 0 \cr
& \Leftrightarrow \sin x.(3.\sin x - 1) = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin x = 0 \hfill \cr
\sin x = {1 \over 3} \hfill \cr} \right. \cr
& \sin x = 0 \Leftrightarrow x = k\pi\,\, (k \in \mathbb Z) \cr
& \sin x = {1 \over 3} \Leftrightarrow \left[ \matrix{
x = \arcsin {1 \over 3} + k2\pi \hfill \cr
x = \pi - {\rm{arcsin}}{1 \over 3} + k2\pi \hfill \cr} \right. \,\,(k \in \mathbb Z)\cr}\)
c) Tìm \(\mathop {\lim }\limits_{x \to 0} {{f''(\sin 5x) + 1} \over {g'(\sin 3x) + 3}}\)
Ta có:
\(f’'(x) = 6x – 1 ⇒ f’’ (sin 5x) = 6.sin5x – 1\)
\(g’(x) = 2x – 3 ⇒ g’(sin 3x) = 2.sin 3x – 3\)
Vậy:
\(\eqalign{
& {{f''(\sin 5x) + 1} \over {g'(\sin 3x) + 3}} = {{6.\sin 5x} \over {2.\sin 3x}} = 5.{{\sin 5x} \over {5x}}.{{3x} \over {\sin 3x}} \cr
& \Rightarrow \mathop {\lim }\limits_{x \to 0} {{f''(\sin 5x) + 1} \over {g'(\sin 3x) + 3}} \cr
& = 5.\mathop {\lim }\limits_{x \to 0} {{\sin 5x} \over {5x}}.\lim {{3x} \over {\sin 3x}} = 5.1.1 = 5 \cr} \)
