Question

Помогите пожалуйста ​

Answer 1

f'(x)=(5x-5x+6)/x²=6/x²

f'(-√6)=6/(-√6)²=1

tgα=1

α=45°

Answer 2

Ответ:   $\varphi =45^\circ$  .

$f(x)=\dfrac{5x-6}{x}\ \ ,\ \ x_0=-\sqrt6\\\\k=f'(x_0)=tg\varphi \\\\f'(x)=\Big(\dfrac{5x-6}{x}\Big)'=\Big(5-\dfrac{6}{x}\Big)'=0-\dfrac{-6}{x^2}=\dfrac{6}{x^2}\\\\\\f'(-\sqrt6)=\dfrac{6}{(-\sqrt6)^2}=\dfrac{6}{6}=1\\\\\\k=tg\varphi =1\ \ ,\ \ \varphi =45^\circ$