Stationary point

A point is stationary for a differentiable function or functional when its first derivative or first variation vanishes in every admissible direction.

The gradient vanishes only when $y=0$ and $\tan x=-2x$. On each tangent branch, $\tan x+2x$ increases strictly from negative to positive infinity, so it has exactly one root $x_s$. With $z_s=e^{-x_s^2}\cos x_s$, the Hessian matrix is diagonal with entries $-(4x_s^2+3)z_s$ and $-2z_s$. Its determinant is positive, so every stationary point is a strict maximum when $z_s>0$ or a strict minimum when $z_s<0$. Signs alternate between successive tangent branches. This gives infinitely many nondegenerate extrema under a decaying Gaussian function envelope.

Where $q(x)\ne 0$, the product rule and chain rule give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q^{\prime }=0$. The denominator uses the square of the real cube root, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.

Suppose $h$ is differentiable and $F^{\prime }>0$ on its range. The multivariable chain rule shows that $F\circ h$ has exactly the same stationary points as $h$. Since $F$ is strictly increasing, the type of every local maximum, local minimum, and saddle point is preserved. If both functions are twice differentiable, then at a stationary pointso the Hessian matrix signature is unchanged. The hypothesis $F^{\prime }>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero derivative and create new stationary points in a composition.