Tentukan fungsi naik dan turun interval f(x)=1/3x²-2x²+3x-6
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[tex]\tt f(x) = \frac{1}{3}x^2-2x^2+3x-6\\ \\f'(x) = \frac{1}{3}(2)x^{2-1}-2(2)x^{2-1} +3\\\\~~~~~~= \frac{2}{3}x-4x+3\\ \\~~~~~~=(\frac{2}{3}-4)x+3\\ \\ ~~~~~=\frac{2-12}{3} x+3\\\\~~~~~=-\frac{10}{3}x+3\\ \\\\Fungsi~naik,~apabila~f'(x) > 0\\\\f'(x) > 0\\\\-\frac{10}{3}x+3 > 0\\ \\-\frac{10}{3}x > -3\\ \\x < \frac{3}{\frac{10}{3} }\\ \\x < \frac{9}{10} \\\\Jadi, fungsi~naik~pada~interval:x < \frac{9}{10}\\[/tex]
[tex]\tt Fungsi~turun,~apabila~f'(x) < 0\\\\f'(x) < 0\\\\-\frac{10}{3}x+3 < 0\\ \\-\frac{10}{3}x < -3\\\\x > \frac{9}{10}[/tex]
Jadi, fungsi turun pada interval : x > ⁹/₁₀
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