Câu hỏi 8 - Mục Bài tập trang 45

1. Nội dung câu hỏi

Cho hàm số \(f\left( x \right) = \sqrt {{x^2} - 2x + 8} \). Giải phương trình \(f'\left( x \right) =  - \frac{2}{3}\).

3. Lời giải chi tiết 

\(f'\left( x \right) \) \( = {\left( {\sqrt {{x^2} - 2x + 8} } \right)'} \) \( = \frac{{\left( {{x^2} - 2x + 8} \right)'}}{{2\sqrt {{x^2} - 2x + 8} }} \) \( = \frac{{2x - 2}}{{2\sqrt {{x^2} - 2x + 8} }} \) \( = \frac{{x - 1}}{{\sqrt {{x^2} - 2x + 8} }}\)

\(f'\left( x \right) = - \frac{2}{3}\) thì \(\frac{{x - 1}}{{\sqrt {{x^2} - 2x + 8} }} = - \frac{2}{3} \) \( \Leftrightarrow \frac{{x - 1}}{{\sqrt {{x^2} - 2x + 8} }} + \frac{2}{3} = 0\)

\( \Leftrightarrow \frac{{3\left( {x - 1} \right) + 2\sqrt {{x^2} - 2x + 8} }}{{3\sqrt {{x^2} - 2x + 8} }} = 0 \) \( \Leftrightarrow 3\left( {x - 1} \right) + 2\sqrt {{x^2} - 2x + 8} = 0\)

\( \Leftrightarrow 2\sqrt {{x^2} - 2x + 8} = - 3\left( {x - 1} \right) \) \( \Leftrightarrow \left\{ \begin{array}{l}x < 1\\4\left( {{x^2} - 2x + 8} \right) = 9{\left( {x - 1} \right)^2}\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}x < 1\\4{x^2} - 8x + 32 = 9{x^2} - 18x + 9\end{array} \right. \) \( \Leftrightarrow \left\{ \begin{array}{l}x < 1\\5{x^2} - 10x - 23 = 0\end{array} \right. \) \( \Leftrightarrow \left\{ \begin{array}{l}x < 1\\\left[ \begin{array}{l}x = \frac{{5 + 2\sqrt {35} }}{5}\\x = \frac{{5 - 2\sqrt {35} }}{5}\end{array} \right.\end{array} \right.\)\( \Leftrightarrow x = \frac{{5 - 2\sqrt {35} }}{5}\)

Vậy \(x = \frac{{5 - 2\sqrt {35} }}{5}\).