**Fermat's Theorem:**
If \(f\) has local max or min at \(c\), and if \(f'(c)\) exists, then \(f'(c)=0.\)

**The converse is not true.**
ATTENTION: \(f'(c)=0\) does not mean \(f(c)\) is a local max or min.

**Counter example.** \(f(x)=x^3\). \(f'(0)=0\) but \(f\) has no max or min (local or global).

**Example of Fermat's Theorem.**
\(f(x)=\sin x\) has a local max at \(x=\frac{\pi}{2}\) and \(f'(\frac{\pi}{2})=0\).

A **critical number** of a function is a number in the domain of \(f\) such that either \(f'(c)=0\) or \(f'(c)\) does not exist.

**Example 1.** Find the critical numbers of \(f(x)=6x^3-9x^2-36x.\)

Sol: Solving \(f'(x)=0\) yields \(x=-1\) and \(x=2\). So \(-1, 2\) are critical numbers.

**Example 2.** Find the critical numbers of \(f(x)=\sqrt{x}.\)

Sol: Domain of \(f\): \(x\geq 0\).
Derivative is \(f'(x)=\frac{1}{2}x^{-1/2}\).
Note that \(f'(x)\neq 0\) for \(x>0\) and \(f'(0)\) does not exist.
Hence the only critical number is \(0\).

(Comment: this is an example where \(0\) is in the domain of \(f\) and \(f'(0)\) does not exist)

**The Closed Interval Method** for finding **global max & min** of **continuous** \(f\) over \([a,b]\):

1. Find values of \(f\) at endpoints of the interval: \(f(a)\), \(f(b)\)

2. Find values of \(f\) at all critical points: first find all critical points, then evaluate \(f\) at those pts

3. Choose the largest from Steps 1,2 as max and the smallest as min

**Example.** Find the global max and min of
\[f(x)=x^3-3x^2+1, \quad -\frac{1}{2}\leq x\leq 4\]

Sol:

**Step 1.** Find values of \(f\) at endpoints:

endpts | \(f\) values |
---|---|

\(\frac{1}{2}\) | \(\frac{1}{8}\) |

\(4\) | \(17\) |

**Step 2.** Find all critical pts and values of \(f\) at those pts. Since \(f'(x)\) exists for all \(x\), all critical pts are roots of
\[f'(x)=3x^2-6x=0.\]
Hence the critical numbers are \(x=0\) and \(x=2\).

Values of \(f\) at critical pts are

critical pts | \(f\) values |
---|---|

\(0\) | \(1\) |

\(2\) | \(-3\) |

**Step 3.**
Candidates of extreme values are \(\frac{1}{8}, 17, 1, -3\).
We see that: max = \(17\), min = \(-3\).

- If \(f'(x)>0\) on an interval, then \(f\) is increasing on that interval
- If \(f'(x)<0\) on an interval, then \(f\) is decreasing on that interval

**Example.** Find where the function
\(f(x)=3x^4-4x^3-12x^2+5\) is increasing or decreasing.

Sol:
\[f'(x)=12x^3-12x^2-24x=12x(x-2)(x+1).\]

**Ex.**
Find the local max and min of \(f(x)=3x^4-4x^3-12x^2+5\).

**concave upward (CU)** and
**concave downward (CD)**

If \(f''(c)=0\), then \(c\) is called an **inflection point**

- \(f''(c)>0\) ==> concave upward ==> \(f\) has a local min at \(c\)
- \(f''(c)<0\) ==> concave downward ==> \(f\) has a local max at \(c\)

**Example.** \(f''\) is continuous near \(c\). \(f'(1)=0\) and \(f''(1)=-1\). What can you say about \(f\) at \(x=1\) ?

Sol: \(f(1)\) is a local maximum.

**Ex.**
Discuss the curve \(y=x^3\) in terms of monotonicity, concavity, inflection point, max/min, etc.