2# uses the second derivative criterion to calculate the local maxima and minima of the following functions: f(x) =x4 -8x²+3

For the function f(x) = x^4 - 8x^2 + 3 , we will determine the local maxima and minima using the second derivative test:

1. Find the first derivative f'(x) by differentiating f(x) :
f'(x) = \frac{d}{dx}(x^4 - 8x^2 + 3) = 4x^3 - 16x .

To find critical points, set f'(x) = 0 :
4x^3 - 16x = 0
4x(x^2 - 4) = 0
4x(x - 2)(x + 2) = 0

This gives critical points x = -2, 0, 2 .

2. Find the second derivative by differentiating f'(x) :
f''(x) = \frac{d}{dx}(4x^3 - 16x) = 12x^2 - 16 .

3. Evaluate f''(x) at each critical point:
- At x = -2 , f''(-2) = 12(-2)^2 - 16 = 32 > 0 , indicating a local minimum at x = -2 .
- At x = 0 , f''(0) = 12(0)^2 - 16 = -16 < 0 , indicating a local maximum at x = 0 .
- At x = 2 , f''(2) = 12(2)^2 - 16 = 32 > 0 , indicating a local minimum at x = 2 .

Therefore, f(x) = x^4 - 8x^2 + 3 has a local maximum at x = 0 and local minima at x = -2 and x = 2 .

\boxed{\text{Answer}}
The function has a local maximum at x = 0 and local minima at x = -2 and x = 2 .