# 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:

- 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).
- 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.