Key takeaway: The extreme value theorem guarantees that a function continuous on a closed interval reaches both an absolute maximum value and an absolute minimum value somewhere on that interval.
Grade 12 – AP Calculus ABAP Calculus AB Unit 5Calculus
Definition of Extreme Value Theorem
The extreme value theorem (EVT) says a continuous function cannot wander off forever or stop just short of a highest or lowest value when it is confined to a closed interval. Picture a hiking trail drawn without lifting the pencil, between two fixed trailheads: somewhere along it there is a highest point and a lowest point.
Those extremes occur either at endpoints or at interior points where the derivative is zero or undefined. That fact turns the theorem into a practical search method for absolute extrema.
Formula and notation
Let \(w\) be continuous on the closed interval \([s,e]\). Then there are inputs \(p\) and \(q\) in \([s,e]\) such that
\[ w(p)\le w(x)\le w(q)\quad\text{for every }x\text{ in }[s,e]. \]
The number \(w(q)\) is the absolute maximum and \(w(p)\) is the absolute minimum.
Key facts
- Both conditions are essential: the function must be continuous, and the interval must be closed.
- On an open interval, a continuous function may have no maximum or minimum at all.
- Candidates for extremes are the endpoints and the critical points inside the interval.
- The theorem guarantees existence only; locating the extremes still needs the candidate test.
Extreme Value Theorem example
Find the absolute maximum and minimum of \(f(x)=2x^3-15x^2+24x+3\) on \([0,6]\).
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The function is a polynomial, so it is continuous, and the interval is closed. The theorem applies.
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Solve \(f^{\prime}(x)=6x^2-30x+24=6(x-1)(x-4)=0\): the critical points are \(x=1\) and \(x=4\).
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Compare values: \(f(0)=3\), \(f(1)=14\), \(f(4)=128-240+96+3=-13\), \(f(6)=432-540+144+3=39\).
The absolute maximum is 39 at \(x=6\), and the absolute minimum is \(-13\) at \(x=4\).
Common mistake
Watch out
A student says \(y=x^2\) is continuous on the open interval \((0,3)\), so by the theorem it must have an absolute minimum there.
Why it is wrong
The interval is open, so neither endpoint belongs to it. The outputs get as close to 0 as you like near \(x=0\), yet never equal 0 inside the interval, and no smallest value exists. Continuity alone is not enough; the extreme value theorem needs a closed, bounded interval.
Related terms and lessons
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