Question

алгебра, 10 класс, с подробным решением​

Answer 1

Ответ:

3.

$f'(x) = \frac{1}{5} \times 5 {x}^{4} - \frac{3}{4} \times 4 {x}^{3} + 3 {x}^{2} - \frac{9}{2} \times 2x = \\ = {x}^{4} - 3 {x}^{3} + 3 {x}^{2} - 9x \\ \\ f'(x) = 0 \\ {x}^{4} - 3 {x}^{3} + 3 {x}^{2} - 9x = 0\\ {x}^{3} (x - 3) + 3x(x - 3) = 0 \\ (x - 3)( {x}^{3} + 3) = 0 \\ x(x - 3)( {x}^{2} + 3) = 0 \\ \\ x_1 = 0 \\ x_2 = 3$

4.

$y = \sqrt{2x + 1} \\ x_0 = 4$

$f(x) = y(x_0) + y'(x_0) + (x - x_0)$

$y(4) = \sqrt{9} = 3$

$y '= \frac{1}{2} {(2x + 1)}^{ - \frac{1}{2} } \times (2x + 1) '= \\ = \frac{2}{2 \sqrt{2x + 1} } = \frac{1}{ \sqrt{2x + 1} }$

$y'(4) = \frac{1}{ \sqrt{9} } = \frac{1}{3}$

$f(x) = 3 + \frac{1}{3} ( x - 4) = 3 + \frac{x}{3} - \frac{4}{3} = \\ = \frac{x}{3} + \frac{5}{3}$

- уравнение касательной

Answer 2

Ответ:

Объяснение:

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