Question

кр по алгебре 10 клас очень ​

Answer 1

1.

а

$f'(x) = 5 {x}^{4} - 12 {x}^{3} + 2$

б

$f'(x) = \frac{(x + 3)'(2x - 5) - (2x - 5)'(x + 3)}{ {(2x - 5)}^{2} } = \\ = \frac{2x - 5 - 2(x + 3)}{ {(2x - 5)}^{2} } = \frac{2x - 5 - 2x - 6}{ {(2x - 5)}^{2} } = \\ = - \frac{7}{ {(2x - 5)}^{2} }$

в

$f'(x) = (( {x}^{2} + 3x) \sqrt{x} ) '= ( {x}^{2} \sqrt{x} + 3x \sqrt{x} ) '= \\ = ( {x}^{ \frac{5}{2} } + 3 {x}^{ \frac{3}{2} } ) '= \frac{5}{2} {x}^{ \frac{3}{2} } + 3 \times \frac{3}{2} {x}^{ \frac{1}{2} } = \\ = 2.5x \sqrt{x} + 4.5 \sqrt{x}$

2.

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

$f'(1) = 1 - 1 = 0 \\ f'(9) = 1 - \frac{1}{3} = \frac{2}{3}$

$f'(x) = 1 - \frac{1}{ \sqrt{x} } \\ f'(x + 1) = 1 - \frac{1}{ \sqrt{x + 1} }$

3.

$f'(x) = 3 \times 3 {x}^{2} - 1 = 9 {x}^{2} - 1 \\ \\ f'(x) = 0 \\ 9 {x}^{2} - 1 = 0 \\ {x}^{2} = \frac{1}{9} \\ x = \pm \frac{1}{3}$

4

$f'(x) = \frac{( {x}^{2} - 8)'(x + 1) - (x + 1)'( {x}^{2} - 8)}{ {(x + 1)}^{2} } = \\ = \frac{2x(x + 1) - ( {x}^{2} - 8) }{ {(x + 1)}^{2} } = \frac{2 {x}^{2} + 2x - {x}^{2} + 8 }{ {(x + 1)}^{2} } = \\ = \frac{ {x}^{2} + 2x + 8}{ {(x + 1)}^{2} } \\ \\ f'(x) 0 \\ \frac{ {x}^{2} + 2x + 8 }{ {(x + 1)}^{2} } 0 \\ \\ {x}^{2} + 2x + 8 = \\ D= 4 - 32 < 0 \\ \\ \frac{ {x}^{2} + 2x + 8}{ {(x + 1)}^{2} } 0 \\ + \: \: \: \: \: \: \: \: \: \: + \\ - - - ( - 1) - - \\ x\in( - \infty ; - 1)U( - 1 ;+ \infty )$

5

$f'(x) = k \\ \\ f'(x) = \cos(x) \\ k = f'( \frac{\pi}{6} ) = \cos( \frac{\pi}{6} ) = \frac{ \sqrt{3} }{2}$