Minima and Maxima

A maximum is a greatest function value, and a minimum is a least function value. Local extrema compare values near a point; absolute, or global, extrema compare values across the entire domain under consideration.

The plural forms are maxima and minima. Together, maxima and minima are called extrema.

Local and absolute extrema

A function has a local minimum at \(x=c\) if \(f(c)\le f(x)\) for all domain points sufficiently close to c. Reverse the inequality for a local maximum.

For an absolute minimum, the inequality must hold for every point in the domain. An absolute maximum is defined similarly. A function need not have either: \(f(x)=x\) has no greatest or least value on the whole real line.

The domain matters. On the closed interval \([0,1]\), that same function has an absolute minimum of 0 and an absolute maximum of 1. On the open interval \((0,1)\), neither value is attained, so neither extremum exists.

Where to look: critical points and endpoints

A critical number is a number c in the domain where \(f'(c)=0\) or \(f'(c)\) does not exist. When a function is differentiable at an interior local extremum, its derivative there must be zero. The converse is false: a zero derivative does not guarantee an extremum.

Also check places where the derivative is undefined. For example, \(f(x)=|x|\) has an absolute minimum at zero, although it has a corner there and \(f'(0)\) does not exist.

The derivative measures the instantaneous rate of change of a function. It is not automatically zero simply because it is evaluated at a single point.

The first derivative test

Inspect the sign of \(f’\) on either side of a critical number:

Positive to negative: the function changes from increasing to decreasing, giving a local maximum.
Negative to positive: it changes from decreasing to increasing, giving a local minimum.
No sign change: there is no local extremum under this test.

For \(f(x)=x^3-3x\),

\[f'(x)=3x^2-3=3(x-1)(x+1).\]

The critical numbers are \(-1\) and \(1\). The derivative is positive before \(-1\), negative between \(-1\) and \(1\), and positive after \(1\). Therefore,

\((-1,2)\) is a local maximum point.
\((1,-2)\) is a local minimum point.

Neither is an absolute extremum on the whole real line, because this cubic is unbounded above and below.

The second derivative test

Suppose \(f'(c)=0\) and \(f\) is twice continuously differentiable near c.

If \(f”(c)>0\), there is a local minimum at c.
If \(f”(c)<0\), there is a local maximum at c.
If \(f”(c)=0\), the test is inconclusive.

In the cubic example, \(f”(x)=6x\). Its values at \(-1\) and \(1\) confirm the maximum and minimum already found.

Why a zero second derivative is inconclusive

At zero, each of the following functions has both first and second derivatives equal to zero:

\(f(x)=x^4\) has a minimum.
\(g(x)=-x^4\) has a maximum.
\(h(x)=x^3\) has neither a maximum nor a minimum; it has a stationary inflection point.

Use the first derivative test or compare nearby function values when the second derivative test fails. A zero second derivative alone does not identify a saddle point.

Finding absolute extrema on a closed interval

A continuous function on a closed, bounded interval attains an absolute maximum and an absolute minimum. To find them, evaluate the function at every interior critical number and at both endpoints.

For example, find the absolute extrema of \(f(x)=x^2-4x+5\) on \([0,3]\).

Find the derivative: \(f'(x)=2x-4\).
Solve \(f'(x)=0\): the only critical number is \(x=2\).
Compare \(f(0)=5\), \(f(2)=1\), and \(f(3)=2\).

The absolute minimum is 1 at \(x=2\), and the absolute maximum is 5 at \(x=0\). Checking only where the derivative is zero would miss the maximum at the endpoint.