Sorry if the formatting is slightly off - this is my first post here.
We are given: $$y = (x^2-1)^4$$
I have found:
$$\frac{d^2y}{dx^2} = 8(x^2-1)^2(7x^2-1)$$
We have to state the set of values of $x$ for which $y = (x^2 - 1)^4$ is concave downwards.
I know that the curve is concave down when $d^2y/dx^2 < 0$, however, in the solutions, it writes that it is concave down when $(7x^2-1) < 0$
This leads to answers of $-\frac{1}{\sqrt7} < x < \frac{1}{\sqrt7}$.
Why is this the case? Any help would be appreciated.
Note that $ (x^2-1)^2 $ is non-negative for all $x \in \mathbb{R}$, so the only chance for $$ y''(x) = 8(x^2-1)^2(7x^2-1) $$ to be negative is if $7x^2-1 < 0$, which you can solve e.g. graphically or by making a sign table.