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Câu hỏi: Cho
\(\eqalign{
& f\left(x \right) = 2{x^3} - {x^2} + \sqrt 3 ; \cr
& g\left(x \right) = {x^3} + {{{x^2}} \over 2} - \sqrt 3 . \cr} \)
Giải bất phương trình \(f'(x) > g'\left(x \right).\)
\(\eqalign{
& f\left(x \right) = 2{x^3} - {x^2} + \sqrt 3 ; \cr
& g\left(x \right) = {x^3} + {{{x^2}} \over 2} - \sqrt 3 . \cr} \)
Giải bất phương trình \(f'(x) > g'\left(x \right).\)
Lời giải chi tiết
\(\begin{array}{l}
f'\left(x \right) = \left({2{x^3} - {x^2} + \sqrt 3 } \right)'\\
= 2.3{x^2} - 2x + 0 = 6{x^2} - 2x\\
g'\left(x \right) = \left({{x^3} + \frac{{{x^2}}}{2} - \sqrt 3 } \right)'\\
= 3.{x^2} + \frac{{2x}}{2} - 0 = 3{x^2} + x\\
f'\left(x \right) > g'\left(x \right) \Leftrightarrow 6{x^2} - 2x > 3{x^2} + x\\
\Leftrightarrow 3{x^2} - 3x > 0 \Leftrightarrow 3x\left({x - 1} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x < 0
\end{array} \right.
\end{array}\)
Vậy \(S=\left( { - \infty; 0} \right) \cup \left({1; + \infty } \right).\)
\(\begin{array}{l}
f'\left(x \right) = \left({2{x^3} - {x^2} + \sqrt 3 } \right)'\\
= 2.3{x^2} - 2x + 0 = 6{x^2} - 2x\\
g'\left(x \right) = \left({{x^3} + \frac{{{x^2}}}{2} - \sqrt 3 } \right)'\\
= 3.{x^2} + \frac{{2x}}{2} - 0 = 3{x^2} + x\\
f'\left(x \right) > g'\left(x \right) \Leftrightarrow 6{x^2} - 2x > 3{x^2} + x\\
\Leftrightarrow 3{x^2} - 3x > 0 \Leftrightarrow 3x\left({x - 1} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
x > 1\\
x < 0
\end{array} \right.
\end{array}\)
Vậy \(S=\left( { - \infty; 0} \right) \cup \left({1; + \infty } \right).\)