Question

Let y=x^2+x, the minimum value of y is

Answer 1

$y = {x}^{2} + x \\ then \: \frac{dy}{dx} = 2x + 1 \\ \frac{ {d}^{2} y}{d {x}^{2} } = 2 \\ hence \: \\ \frac{dy}{dx} = 0 \: for \: min \\ 2x + 1 = 0 \\ x = \frac{ -1 }{2} \\ putting \: x = \frac{ - 1}{2} \\ we \: get \\ y = \frac{1}{4} - \frac{1}{2} = \frac{ - 1}{4}$

Answer 2

Answer:

Hey, see the attachment. Cheers!