Extreme values and critical numbers
“Nature is thrifty in all its actions.”
Largest compared with what?
The top of a small hill is higher than nearby ground, but it need not be the highest point in a landscape. Calculus makes the same distinction between a local comparison and an absolute one. The domain is part of the question: restricting the landscape can change its highest and lowest points.
Definition
Local and absolute extrema
A function \(f\) has an absolute maximum at \(c\) on its stated domain \(D\) if \(f(x)\le f(c)\) for every \(x\in D\). Reverse the inequality for an absolute minimum.
It has a local maximum at an interior input \(c\) if the same comparison holds for all domain inputs sufficiently close to \(c\). Reverse the inequality for a local minimum. Together, maxima and minima are called extrema.
We use the interior-point convention for local extrema in this course. Endpoints still matter for absolute extrema. The input \(c\) is the location; \(f(c)\) is the extremum value.
Theorem
Extreme-value theorem
If \(f\) is continuous on a closed bounded interval \([a,b]\), then \(f\) attains both an absolute maximum and an absolute minimum there.
The theorem guarantees that both values are achieved. It does not say they are unique, that they occur in the interior, or how to locate them. We state this existence theorem without proof.
Each hypothesis matters. Choose a case below to see how the conclusion can fail.
Why an interior smooth extremum has zero slope
At a smooth local peak, the graph rises toward the point and falls away. We can make the horizontal-tangent conclusion precise directly from the derivative definition.
Theorem
Fermat's theorem
If \(f\) has a local extremum at an interior input \(c\) and is differentiable there, then \(f^{\prime}(c)=0\).
Proof (optional)
At a local maximum, \(f(c+h)-f(c)\le0\) for sufficiently small \(h\). For positive \(h\) the difference quotient is nonpositive; for negative \(h\) it is nonnegative. If the common derivative limit exists, it must be both nonpositive and nonnegative, hence zero. At a local minimum the inequalities reverse and yield the same conclusion.
There are three ways a candidate can escape the horizontal-tangent picture: it can be an endpoint, a point where the derivative does not exist, or a horizontal tangent that is not an extremum at all.
Definition
Critical numbers
An interior input \(c\) in the domain of \(f\) is a critical number if \(f^{\prime}(c)=0\) or \(f^{\prime}(c)\) does not exist as a finite number. The point \((c,f(c))\) is a critical point of the graph.
A missing domain input is not a critical number. Every interior local extremum is critical, but the converse need not hold.
Example 1
Three horizontal or sharp behaviours
The function \(x^2\) has derivative zero and a local minimum at zero. The function \(x^3\) also has derivative zero there, but takes smaller values to the left and larger values to the right: no local extremum occurs. The function \(\lvert x\rvert\) has no derivative at zero, yet has a minimum there.
So “solve \(f^{\prime}=0\)” is only part of a search. It misses corners and cannot classify the solutions on its own.
Find absolute extrema on a closed interval
If \(f\) is continuous on \([a,b]\), existence is guaranteed. Fermat’s theorem says that every interior extremum is a critical number. Therefore:
- Find the critical numbers in \((a,b)\), including points with no finite derivative.
- Evaluate \(f\) at those inputs and at both endpoints.
- Compare the function values, and report all inputs where the largest and smallest values occur.
The derivative locates candidates. The original function decides their heights.
Example 2
An interior peak and a larger endpoint
Find the absolute extrema of \(f(x)=x^3-3x\) on \([-2,3]\). The function is continuous. Its derivative is
\[\begin{aligned} f^{\prime}(x)&=3(x^2-1)\\ &=3(x-1)(x+1). \end{aligned}\]The interior critical numbers are \(-1\) and \(1\). Compare all four candidates:
| Input \(x\) | \(-2\) | \(-1\) | \(1\) | \(3\) |
|---|---|---|---|---|
| Value \(f(x)\) | \(-2\) | \(2\) | \(-2\) | \(18\) |
The absolute maximum value is \(18\) at \(x=3\). The absolute minimum value is \(-2\) at both \(x=-2\) and \(x=1\). The interior peak at \(x=-1\) is only local.
Example 3
A candidate where differentiation fails
Find the absolute extrema of \(f(x)=\lvert x-1\rvert\) on \([-1,4]\). It is continuous. Its derivative is \(-1\) for \(x<1\) and \(1\) for \(x>1\); no derivative exists at \(1\).
The candidates are \(-1,1,4\), with values \(2,0,3\). Thus the absolute minimum is \(0\) at \(1\) and the absolute maximum is \(3\) at \(4\). There were no solutions to \(f^{\prime}(x)=0\), but there was an interior minimum.
Attention
When the domain changes
For \(f(x)=x^2\) on \((0,2]\), the maximum \(4\) is attained at \(2\), but there is no minimum: zero is approached and never reached. A boundary limit is not an attained extremum. On an open or unbounded domain, inspect what happens toward its boundaries instead of invoking the extreme-value theorem automatically.