Question

Срочно!!Пожалуйста, 40б

Answer 1

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Answer 2

$f(x) = \frac{ x^{3} }{3} -5 x^{2} +28x-7\\\\f'(x)= x^{2} -10x+28\\\\f'( x_{0})= x_{0} ^{2}-10 x_{0}+28\\\\ x_{0} ^{2} } -10 x_{0}+28=3\\\\ x_{0} ^{2}-10 x_{0}+25=0\\\\( x_{0}-5) ^{2} =0\\\\ x_{0}-5=0\\\\ x_{0} =5$


$y_{0}= \frac{ 5^{3} }{3}-5*5 ^{2} +28*5-7= \frac{125}{3} -125+140-7=41 \frac{2}{3} +8=49,6$

Ответ: (5 , 49,6)