f(x) = x^4 - 4x^2 - 3 is concave down on the interval I = (Approximate your answer to 3 decimal places.)

Thanks, Anonymous....you are correct......I forgot about that!!!

The correct interval is (-√(2/3), √(2/3))

Let me add one proviso to what Anonymous said.......setting the second derivative to 0 and solving only indicates possible  inflection points. For instance...the second derivative of x^4 is 12x^2. Setting this to 0 gives us that x = 0. But this isn't an inflection point because this curve is always concave "up". To see this, pick a point on either side of 0 and "plug" them into the second derivative. Both results will be positive....indicating that the curve is concave "up" on both sides. Thus, there is no inflection point !!!

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

Concavity changes at points of inflexion. To find such  points, equate the second derivative to zero.

I think that leads to $$x=\pm\sqrt{2/3}.$$