Задача 53628 ...

5f5f83f2116bb85615e04ac9

24 сентября 2020 г. в 11:34

45.9. Пусть f(x) = (5x + 4)¹³ / (x – 3)⁶. Решите неравенство:

1) f(x) > 0; 2) f(x) ≥ 0; 3) f(x) < 0; 4) f(x) ≤ 0.

5f3ea7e3faf909182968ddd9

24 сентября 2020 г. в 12:26

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По формуле производная частного.

[m](\frac{u}{v})`=\frac{u`\cdot v-u\cdot v`}{u^2}[/m]

[m]f`(x)=\frac{13\cdot (5x+4)^{12}\cdot (x-3)^{6}-(5x+4)^{13}\cdot 6\cdot (x-3)^5}{((x-3)^6)^2}[/m]

[m]f`(x)=\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(13 \cdot (x-3)-(5x+4)\cdot 6)}{(x-3)^{12}}[/m]

[m]f`(x)=\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(13x-39-30x-24)}{(x-3)^{12}}[/m]

[m]f`(x)=\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(-17x-63)}{(x-3)^{12}}[/m]

Решаем методом интервалов;

1) [m] f`(x) >0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(-17x-63)}{(x-3)^{12}}>0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(17x+63)}{(x-3)^{12}}<0[/m]


__–___ (–63/17) ___+__ (–4/5) ___–___ (–3) __+___

О т в е т. [m](-\infty; -\frac{63}{17}) \cup (-\frac{4}{5}; -3)[/m]


2) [m] f`(x) \geq 0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(-17x-63)}{(x-3)^{12}}\geq0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(17x+63)}{(x-3)^{12}}\leq0[/m]


__–___ [–63/17] ___+__ [–4/5] ___–___ (3) __+___

О т в е т. [m](-\infty; -\frac{63}{17}] \cup [-\frac{4}{5}; 3)[/m]

3) [m] f`(x) <0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(-17x-63)}{(x-3)^{12}}<0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(17x+63)}{(x-3)^{12}}>0[/m]


__–___ (–63/17) ___+__ (–4/5) ___–___ (3) __+___

О т в е т. [m](-\frac{63}{17};-\frac{4}{5})\cup(3;+\infty)[/m]


4) [m] f`(x) \leq 0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(-17x-63)}{(x-3)^{12}}\leq0[/m]

[m]\frac{(5x+4)^{12}\cdot (x-3)^{5}\cdot(17x+63)}{(x-3)^{12}}\geq0[/m]


__–___ [–63/17] ___+__ [–4/5] ___–___ (3) __+___

О т в е т. [m][-\frac{63}{17};-\frac{4}{5}]\cup(3;+\infty)[/m]