Question

3-A função y=x2-2x+1 tem minimo no ponto em
que x vale:
a)
b)1
12
d)3​

Answer 1

$\sf y = x {}^{2} - 2x + 1 \\ \\ \\ \sf 0 = x {}^{2} - 2x + 1 \\ \\ \\ \sf 0 - x {}^{2} + 2x - 1 = 0 \\ \\ \\ \sf  - x {}^{2} + 2x - 1 = 0 \\ \\ \\ \sf  - x {}^{2} + x + x - 1 = 0 \\ \\ \\ \sf  - x \cdot(x - 1) + x - 1 = 0 \\ \\ \\ \sf  - x \cdot(x - 1) + 1(x - 1) = 0 \\ \\ \\ \sf  - (x - 1) \cdot(x - 1) = 0 \\ \\ \\ \sf  - (x - 1) {}^{1} \cdot(x - 1) = 0 \\ \\ \\ \sf  - (x - 1) {}^{1} \cdot(x - 1) {}^{1} = 0 \\ \\ \\ \sf  - (x - 1) {}^{1 + 1} = 0 \\ \\ \\ \sf  - (x - 1) {}^{2} = 0 \cdot( - 1) \\ \\ \\ \sf (x - 1) {}^{2} = 0 \\ \\ \\ \sf x - 1 = 0 \\ \\ \\ \sf {\color{red}x = 1}$

Att: Nerd1990