Q.Find the critical numbers (only x values) of the function $f(x)=x^{4/5}(x-4)^2$.

Absolute (global) extrema. For $f$ defined on a domain $D$, $f(x_0)$ is the absolute maximum of $f$ on $D$ if $f(x_0)\ge f(x)$ for every $x\in D$; the absolute minimum is defined symmetrically with $\le$.

Extreme Value Theorem. If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a,b]$ — and the extremum can only occur either at an interior critical number or at one of the two endpoints.

Procedure for absolute extrema on $[a,b]$ (Exercise 7.6 Q1's method):

1. Find every critical number of $f$ in the open interval $(a,b)$.
2. Evaluate $f$ at each critical number and at both endpoints $a,b$.
3. The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Relative (local) extrema. $f$ has a relative (local) maximum at $x_0$ if $f(x_0)$ is the largest value of $f$ on some open interval around $x_0$ (relative minimum: smallest, on some open interval). A function may have several local extrema, and a local extremum need not be the absolute one.

Fermat's Theorem. If $f$ has a relative extremum at $x=c$, then $c$ must be a critical number of $f$ (so the search for local extrema always starts by solving $f'(x)=0$ together with any points where $f'$ fails to exist) — though not every critical number is automatically an extremum (e.g. $y=x^3$ at $x=0$).

First Derivative Test. At a critical point $c$ where $f$ is continuous, examine the sign of $f'(x)$ moving left to right across $c$:

• negative $\to$ positive: local minimum at $c$;
• positive $\to$ negative: local maximum at $c$;
• no sign change (same sign on both sides): $c$ is neither a local max nor a local min.

Second Derivative Test (an alternative, often quicker, at a stationary point). If $f'(c)=0$ and $f''(c)$ exists:

• $f''(c)<0$ $\Rightarrow$ local maximum at $c$;
• $f''(c)>0$ $\Rightarrow$ local minimum at $c$; …