Question

Пооомогите пожалуйста с алгеброй(((((

Answer 1

$\displaystyle\bf\\1)\\\\f(x)=\sqrt{4x+1} \\\\\\f'(x)=\frac{1}{2\sqrt{4x+1} } \cdot(4x+1)'=\frac{1}{2\sqrt{4x+1} } \cdot 4=\frac{2}{\sqrt{4x+1} } \\\\\\Otvet \ : \ f'(x)=\frac{2}{\sqrt{4x+1} } \\\\2)\\\\f(x)=x^{2} -5x\\\\f'(x)=(x^{2} )'-5\cdot x'=2x-5\\\\tg\alpha =tg45^\circ=1\\\\f'(x_{0} )=tg\alpha =1\\\\2x-5=1\\\\2x=6\\\\x=3\\\\Otvet \ : \ B)$