Linearization and constrained optimization

What this concept is

Linearization uses f(a)+f′(a)(x−a) near a. To find extrema on a closed bounded interval for continuous f, compare endpoints with interior critical points where f′=0 or does not exist.

Why it matters here

A local stationary point is not automatically a global optimum. Constraints and units determine which candidate is usable.

Concrete example

Minimize C(x)=x+16/x on [1,8]. C′=1−16/x² vanishes at x=4. Values C(1)=17, C(4)=8, C(8)=10 establish the global minimum among candidates. Also C′ is negative before 4 and positive after it. Approximate sqrt(4.04) by 2+(1/4)·0.04=2.01; concavity predicts this tangent estimate is an overestimate.

Common confusion / misconception

Solving f′=0 alone misses endpoints and corners. A second derivative equal to zero is inconclusive; x³ has no extremum at zero.

How to use it

Define feasible inputs first. Find candidates, evaluate original objective, and check units. The mean value theorem says a function continuous on [a,b] and differentiable inside has some c with f′(c)=[f(b)−f(a)]/(b−a). It explains why a derivative bound controls change. It does not identify c for you.

Check yourself

Where does x² attain its maximum on [−1,2]? Why would a stationary-point-only search fail?

Mini drill or application

Minimize x+9/x on [1,6], and then [4,6]. Answers: x=3 with value 6; x=4 with value 25/4. Explain the changed constraint rather than repeating the unconstrained answer.

Read this only if stuck

OpenStax selective support. Use the reading guide for the section sequence and practice contract. Record your reasoning before consulting solutions.