Question

If y = x² + 4x + 2, find the value of x at which y is minimum.​

Answer 1

Step-by-step explanation:

We have,

$y = {x}^{2} + 4x + 2$

For minimum,

• $\tt\purple{\frac{dy}{dx}=0}$
• $\tt\purple{\frac{d^2y}{dx^2}>0}$

So,

$\frac{dy}{dx} = 2x + 4 = 0$

$\implies 2x = - 4$

$\implies x = - 2$

Now,

$\frac{d^{2} y}{dx^{2} } = 2$

$\implies\frac{d^{2} y}{dx^{2} } \bigg|_{x = - 2} = 2 > 0$

So,

at $x=-2$,

$\sf \pink{y \: \: attains \: \: minimum \: \: value}$