решите уравнение f '(x)=0:
1) f(x)=4x^2+2x
2) f(x)=x^2+x-1
напишите пожалуйста полное решение
1)f(x)=8х+2=0 8х+-2 х=-1/4 2) f(х)= 2х+1=0 2х=-1 х=-1/2 наверное так
1. (4x^2+2x)' = 8x+2
f'(x) = 0; 8x+2 = 0
8x = -2
x = -0.25
2. (x^2+x-1)' = 2x+1
f'(x) = 0; 2x+1 = 0
2x = -1
x = -0.5
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Answers & Comments
1)f(x)=8х+2=0 8х+-2 х=-1/4 2) f(х)= 2х+1=0 2х=-1 х=-1/2 наверное так
1. (4x^2+2x)' = 8x+2
f'(x) = 0; 8x+2 = 0
8x = -2
x = -0.25
2. (x^2+x-1)' = 2x+1
f'(x) = 0; 2x+1 = 0
2x = -1
x = -0.5