Question

y =x^2 +4x find minimum value of y​

Answer 1

GIVEN :-

• y = x² + 4x

TO FIND :-

• Minimum value of y.

TO KNOW :-

Let y = xⁿ.

$\\ \boxed{ \sf \: \dfrac{dy}{dx} = n {x}^{n - 1} }$

dy/dx = 0 , for minimum value of 'y'.

SOLUTION :-

$\\ \sf \: y = {x}^{2} + 4x$

Differentiating with respect to x ,

$\\ \sf \: \dfrac{dy}{dx} = \dfrac{d( {x}^{2} + 4x) }{dx} \\ \\ \sf \dfrac{dy}{dx} = 2x + 4$

For minimum value , dy/dx = 0.

Hence ,

→ 2x + 4 = 0

→ 2x = -4

→ x = -4/2

→ x = -2

Putting value of x in the given equation,

→ y = x² + 4x

→ y = (-2)² + 4(-2)

→ y = 4 - 8

→ y = -4

Hence , minimum value of y is -4.