 Krishna
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Step 1: Find the first derivative of the function

Step 1; Apply the deviation on both both sides

\frac{dy}{dx} = \frac{dy}{dx} (x^4 - 8x^2 + 3)

Step 2: Differentiate the given equation

\frac{dy}{dx} = 4x^3 - 16x

4x^3 - 16x = 0

Step 3: Solve the equation of unknown values

$4x[(x^2) - 4] = 0$

x^2 = 4

x=\pm2

Step 2: Calculate the second derivative of the function

\frac{d^2y}{dx^2} = \frac{d^2y}{dx^2}(4x^3 - 16x)

\frac{d^2y}{dx^2} = 12x^2 - 16

Step 3: Determining the nature of stationary points.

NOTE;  Use your values of x if necessary to check the sign of the second derivative at each of the stationary points.

RULES: If   \frac{d^2y}{dx^2} is positive, then it is a minimum point

If   \frac{d^2y}{dx^2} is negative, then it is a maximum point

If   \frac{d^2y}{dx^2} = zero, then it could be a maximum, minimum or point of inflexion.

[Case 1: Both y-coordinates less than that of the stationary point: maximum

Case 2: Both y-coordinates greater than that of the stationary point: minimum

Case 3: One y-coordinates less than and one greater than that of the

stationary point: Point of inflection ]

EXAMPLE: x = 0,   \frac{d^2y}{dx^2} = 12x^2 - 16

\frac{d^2y}{dx^2}=\ -16 maximum

x = 2,   \frac{d^2y}{dx^2} = (12 * 4) - 16 =  32  minimum

Hence there is a minimum point at x = ± 2 and a maximum point at x = 0 