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.