For this question from Introduction to optimisation by Chong & Zak :
The answer is :
What I'm unsure about is why $h$ and $g$ are not considered to be linearly independent.
Is this because for the points in the feasible region we have
$$ x = [2, \alpha], \;\; \alpha \in [-\infty , -1] $$
Then
$$ \nabla h(x) = [0, 0]^T $$
and
$$ \nabla g(x) = [0, 3(x_2 + 1)^2]^T $$
For the values of $x$ it is possible to choose $x_2 = -1$ then
$$ \nabla h = \nabla g $$
meaning that, even though it isn't generally true, there is some value of $x$ for which we can express $\nabla h$ in terms of $\nabla g$.
If that isn't it then I'm not sure what it is.
In the feasible region, we have $x_1=2$, hence $\nabla h$ is identically $0$.
To verify linear dependence, simply note that $(1)(\nabla h) + (0)(\nabla g) = 0$.