Answer :
To identify the minimum value of the function [tex]\( y = 2x^2 + 4x \)[/tex], we need to follow several steps:
1. Find the first derivative of the function:
To locate the critical points where the minimum might occur, we first take the first derivative of the function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + 4x) = 4x + 4 \][/tex]
2. Set the first derivative equal to zero:
Next, we set the first derivative equal to zero to find the critical points:
[tex]\[ 4x + 4 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 4x = -4 \implies x = -1 \][/tex]
So, the critical point is [tex]\( x = -1 \)[/tex].
3. Determine the nature of the critical point:
We need to find the second derivative of the function to determine whether the critical point is a minimum, maximum, or inflection point:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x + 4) = 4 \][/tex]
Since the second derivative is positive ([tex]\( 4 > 0 \)[/tex]), the function is concave upwards at [tex]\( x = -1 \)[/tex]. This indicates that [tex]\( x = -1 \)[/tex] is a point of local minimum.
4. Evaluate the function at the critical point:
To find the minimum value of the function, we evaluate [tex]\( y \)[/tex] at [tex]\( x = -1 \)[/tex]:
[tex]\[ y(-1) = 2(-1)^2 + 4(-1) = 2(1) - 4 = 2 - 4 = -2 \][/tex]
Therefore, the minimum value of the function [tex]\( y = 2x^2 + 4x \)[/tex] is [tex]\(\boxed{-2}\)[/tex].
1. Find the first derivative of the function:
To locate the critical points where the minimum might occur, we first take the first derivative of the function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + 4x) = 4x + 4 \][/tex]
2. Set the first derivative equal to zero:
Next, we set the first derivative equal to zero to find the critical points:
[tex]\[ 4x + 4 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ 4x = -4 \implies x = -1 \][/tex]
So, the critical point is [tex]\( x = -1 \)[/tex].
3. Determine the nature of the critical point:
We need to find the second derivative of the function to determine whether the critical point is a minimum, maximum, or inflection point:
[tex]\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(4x + 4) = 4 \][/tex]
Since the second derivative is positive ([tex]\( 4 > 0 \)[/tex]), the function is concave upwards at [tex]\( x = -1 \)[/tex]. This indicates that [tex]\( x = -1 \)[/tex] is a point of local minimum.
4. Evaluate the function at the critical point:
To find the minimum value of the function, we evaluate [tex]\( y \)[/tex] at [tex]\( x = -1 \)[/tex]:
[tex]\[ y(-1) = 2(-1)^2 + 4(-1) = 2(1) - 4 = 2 - 4 = -2 \][/tex]
Therefore, the minimum value of the function [tex]\( y = 2x^2 + 4x \)[/tex] is [tex]\(\boxed{-2}\)[/tex].
