Cue Column:

• How to find absolute max/min?
• Why check endpoints?
• What about open intervals?

Notes Column: Extreme Value Theorem: If f is continuous on [a,b], then f attains both an absolute maximum and an absolute minimum on [a,b].

Method (CENT):

1. Find all Critical points in (a,b) where f'(x) = 0 or f'(x) DNE
2. Evaluate f at all critical points
3. Evaluate f at the Endpoints a and b
4. Compare: largest value = absolute max, smallest = absolute min

On open intervals or unbounded domains: Global extrema are NOT guaranteed. Example: f(x) = 1/x on (0, inf) has no maximum and no minimum. For these cases, analyze limits as x approaches boundary points.

JEE Approach: For word problems asking "maximum value" of f on [a,b], always check endpoints! Many students only find critical points and miss that the endpoint value is larger.

Example: f(x) = $x^3$ - 3x + 1 on [-2, 2]. f'(x) = 3$x^{2-3}$ = 0 at x = +/-1. f(-2) = -8+6+1 = -1, f(-1) = -1+3+1 = 3, f(1) = 1-3+1 = -1, f(2) = 8-6+1 = 3. Max = 3 (at x=-1 and x=2), Min = -1 (at x=-2 and x=1).

Summary: For closed intervals, the algorithm is mechanical: find critical points, evaluate at critical points and endpoints, compare.