1. Найдем \( f(9) \):
\( f(x) = (5x-4)\sqrt{x} \)
\( f(9) = (5 \cdot 9 - 4) \sqrt{9} = (45 - 4) \cdot 3 = 41 \cdot 3 = 123 \)
2. Найдем \( f'(x) \):
Сначала преобразуем функцию: \( f(x) = (5x-4)x^{1/2} = 5x^{3/2} - 4x^{1/2} \).
Теперь найдем производную:
\( f'(x) = \frac{d}{dx}(5x^{3/2}) - \frac{d}{dx}(4x^{1/2}) \)
\( f'(x) = 5 \cdot \frac{3}{2}x^{3/2 - 1} - 4 \cdot \frac{1}{2}x^{1/2 - 1} \)
\( f'(x) = \frac{15}{2}x^{1/2} - 2x^{-1/2} \)
\( f'(x) = \frac{15}{2}\sqrt{x} - \frac{2}{\sqrt{x}} \)
3. Найдем \( f'(9) \):
\( f'(9) = \frac{15}{2}\sqrt{9} - \frac{2}{\sqrt{9}} = \frac{15}{2} \cdot 3 - \frac{2}{3} = \frac{45}{2} - \frac{2}{3} = \frac{135 - 4}{6} = \frac{131}{6} \)
4. Найдем \( f(9) - f'(9) \):
\( f(9) - f'(9) = 123 - \frac{131}{6} = \frac{123 \cdot 6 - 131}{6} = \frac{738 - 131}{6} = \frac{607}{6} \)
Ответ: 1) \( f(9) = 123 \); 2) \( f'(x) = \frac{15}{2}\sqrt{x} - \frac{2}{\sqrt{x}} \); 3) \( f'(9) = \frac{131}{6} \); 4) \( f(9) - f'(9) = \frac{607}{6} \).