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Code Scholars Tutoring · AP Calculus AB

Absolute extrema on an interval

Unit 5 · Topics 5.5. Original AP Calculus AB notes, worked example, and self-check.

Understand the idea

On a closed interval where the function is continuous, evaluate the function at every interior critical number and at both endpoints. Compare the actual function values to identify the absolute extrema. Points outside the domain are not candidates. On an open or unbounded domain, analyze endpoint limits and justify whether extrema are attained instead of blindly applying the closed-interval procedure.

Candidates on [a,b]:a, b, {c∈(a,b):f′(c)=0 or undefined}\text{Candidates on }[a,b]:\quad a,\ b,\ \{c\in(a,b):f'(c)=0\text{ or undefined}\}

Worked example

Find the absolute extrema of f(x)=x3−3x\displaystyle f(x)=x^3-3x on [−2,2]\displaystyle [-2,2].

Critical numbers are −1\displaystyle -1 and 1\displaystyle 1. Evaluate f(−2)=−2\displaystyle f(-2)=-2, f(−1)=2\displaystyle f(-1)=2, f(1)=−2\displaystyle f(1)=-2, and f(2)=2\displaystyle f(2)=2. The absolute maximum value is 2\displaystyle 2 at −1\displaystyle -1 and 2\displaystyle 2; the absolute minimum is −2\displaystyle -2 at −2\displaystyle -2 and 1\displaystyle 1.

Try it yourself

Find the absolute minimum of x+4/x\displaystyle x+4/x on [1,4]\displaystyle [1,4].

Show solution and reasoning

The derivative 1−4/x2\displaystyle 1-4/x^2 vanishes at 2\displaystyle 2. Candidate values are 5\displaystyle 5, 4\displaystyle 4, and 5\displaystyle 5, so the absolute minimum is 4\displaystyle 4 at x=2\displaystyle x=2.