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Second Derivative Test

Instead of using First Derivative Test we can use another test.

Second Derivative Test.

  • if $$${f{'}}{\left({c}\right)}={0}$$$ and $$${f{''}}{\left({c}\right)}<{0}$$$ then there is maximum at point $$${c}$$$.
  • if $$${f{'}}{\left({c}\right)}={0}$$$ and $$${f{''}}{\left({c}\right)}>{0}$$$ then there is minimum at point $$${c}$$$.
  • if $$${f{'}}{\left({c}\right)}={0}$$$ and $$${f{''}}{\left({c}\right)}={0}$$$ then we can't say anything about point $$${c}$$$.

This test is used not so often as first derivative test because of two reasons:

  1. We can't apply it to stationary points for which first derivative doesn't exist (because in this case second derivative also doesn't exist).
  2. It is inconclusive when second derivative equals 0.

Example. Find extrema of the function $$${y}={{x}}^{{4}}-{4}{{x}}^{{3}}$$$.

We first find first two derivatives.

$$${f{'}}{\left({x}\right)}={4}{{x}}^{{3}}-{12}{{x}}^{{2}}={4}{{x}}^{{2}}{\left({x}-{3}\right)}$$$.

So, critical numbers (points where $$${f{'}}{\left({x}\right)}={0}$$$) are $$${x}={0},{x}={3}$$$.

$$${f{''}}{\left({x}\right)}={\left({f{'}}{\left({x}\right)}\right)}'={\left({4}{{x}}^{{3}}-{12}{{x}}^{{2}}\right)}'={12}{{x}}^{{2}}-{24}{x}={12}{x}{\left({x}-{2}\right)}$$$.

Since $$${f{''}}{\left({3}\right)}={12}\cdot{3}\cdot{\left({3}-{2}\right)}={36}>{0}$$$ then according to second derivative test $$${x}={3}$$$ is minimum.

Since $$${f{''}}{\left({0}\right)}={12}\cdot{0}{\left({0}-{2}\right)}={0}$$$ then we can't say anything about $$${x}={0}$$$ using second derivative test. Using first derivative test we conclude that $$${x}={0}$$$ is neither minimum nor maximum.