Higher-Order Derivative Test
As we know Second derivative test is inconlusive when for critocal point $$${c}$$$ $$${f{''}}{\left({c}\right)}={0}$$$.
In this case we need to use another test.
Higher-Order Derivative Test. Suppose that $$${k}$$$ $$$\left({k}>{2}\right)$$$ is the smallest number for which $$${{f}}^{{{\left({k}\right)}}}{\left({c}\right)}\ne{0}$$$. If $$${k}$$$ is and odd number, then there is no maximum or minimum at $$${c}$$$. If $$${k}$$$ is even number then $$${c}$$$ is maximum if $$${{f}}^{{{\left({k}\right)}}}{\left({c}\right)}<{0}$$$, and $$${c}$$$ is minimum if $$${{f}}^{{{\left({k}\right)}}}{\left({c}\right)}>{0}$$$.
Example. Find and classify extrema of the function $$${f{{\left({x}\right)}}}={{e}}^{{x}}+{{e}}^{{-{x}}}+{2}{\cos{{\left({x}\right)}}}$$$.
$$${f{'}}{\left({x}\right)}={{e}}^{{{x}}}-{{e}}^{{-{x}}}-{2}{\sin{{\left({x}\right)}}}$$$.
$$${f{'}}{\left({x}\right)}={0}$$$ only when $$${x}={0}$$$.
So, there is only one stationary point $$${x}={0}$$$.
$$${f{''}}{\left({x}\right)}={{e}}^{{x}}+{{e}}^{{-{x}}}-{2}{\cos{{\left({x}\right)}}}$$$. Since $$${f{''}}{\left({0}\right)}={0}$$$ second derivative test is inconclusive.
$$${f{'''}}{\left({x}\right)}={{e}}^{{x}}-{{e}}^{{-{x}}}+{2}{\sin{{\left({x}\right)}}}$$$. Since $$${f{'''}}{\left({0}\right)}={0}$$$ then we can't say anything about point $$${x}={0}$$$.
$$${{f}}^{{{\left({4}\right)}}}{\left({x}\right)}={{e}}^{{x}}+{{e}}^{{-{x}}}+{2}{\cos{{\left({x}\right)}}}$$$. Since $$${{f}}^{{{\left({4}\right)}}}{\left({0}\right)}={4}>{0}$$$ and order of derivative is even number then $$${x}={0}$$$ is minimum according to Higher-Order Derivative Test.
Note, that there are still examples of non-constant functions whose derivatives of all orders at critical point equal 0.