f(x) = x^4 - 4x^2 - 3
f'(x) = 4x^3 - 8x
Setting the first derivative to 0 to find the critical points, we have
4x^3 - 8x = 0 factor
4x(x^2 - 2) = 0 and setting each factor to 0 we have x = 0, x =±√(2)
And finding the second derivative, f"(x), we have .... 12x^2 - 8
And "plugging the critical points into this we have
f"(0) = -8 so the curve is concave down at x = 0
And
f"(±√(2)) = 12(2) - 8 = 24-8 = 16, so the curve is concave up at x= ±√(2)
So, the curve is concave down on (-√2, √(2 ) = (-1.414, 1.414)
Here's the graph ...https://www.desmos.com/calculator/1ffgrl0kji