Calculus Applications of differentiation

Local maxima and minima

Q: How do you find local maxima and minima?

Solve f prime of x equals 0 for critical points, then classify each with the second derivative test or a first derivative sign chart.

Q: What is the second-derivative test?

At a critical point c: if the second derivative is positive the point is a local minimum; if negative it is a local maximum. If zero the test is inconclusive.

Q: What is the first-derivative test?

Check how the derivative changes sign at the critical point: positive to negative is a maximum, negative to positive is a minimum, no change is neither.

Q: What is the difference between the location and the value of a maximum?

The location is the x value c where it occurs; the value is f of c, the height of the curve there.

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Local Maxima and Minima

California Calculus • Standard 9.0 • Applications of Differentiation

Local Maxima and Minima is a topic in Applications of Differentiation in the California Calculus Standards. It is aligned to Standard 9.0, which requires students to use differentiation to sketch graphs by hand, here locating the local maxima and minima of a function.

A local maximum or minimum occurs at a critical point where a curve turns around. Two tests classify it: the first-derivative test (a sign change of \(f'\)) and the second-derivative test (the sign of \(f''\)).

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Theory

A local maximum or minimum occurs at a critical point where the curve turns around. Two tests classify it: the first-derivative test (does \(f'\) change sign?) and the second-derivative test (is \(f''\) positive or negative?).

At a critical point (\(f'=0\)) the curve may turn around. Two tests decide what happens:

First-derivative test: if \(f'\) changes \(+\to-\) it is a local max; \(-\to+\) is a local min; no change is neither.
Second-derivative test: \(f''(c)>0\) means a local min (concave up), \(f''(c)<0\) means a local max (concave down); \(f''(c)=0\) is inconclusive.

The extreme value is \(f(c)\) — substitute the critical \(x\) back into \(f\).

Key idea: the second-derivative test is fastest when \(f''\) is easy; fall back to the first-derivative test when \(f''(c)=0\).

Local max and min: each has a horizontal tangent (\(f'=0\)).

First-derivative test: \(f'\) changes \(+\to-\) at a local max.

The second-derivative test at a critical point \(c\):

\[f''(c)>0\ \Rightarrow\ \text{local min},\qquad f''(c)<0\ \Rightarrow\ \text{local max}\]

f double prime positive gives a minimum, negative gives a maximum

Always report the value. The location is \(x=c\); the local extreme value is \(f(c)\).

How to find and classify local extrema

Find critical points: solve \(f'(x)=0\).
Classify each with the second-derivative test (or a sign chart of \(f'\)).
Evaluate \(f(c)\) for the local maximum or minimum value.

Example 1 — Second-derivative test

Classify the critical point of \(f(x)=x^2-4x+1\).

Solution

Find the critical point, then test with \(f''\).

\(f'(x)\)	\(=\)	\(2x-4=0\Rightarrow x=2\)
\(f''(x)\)	\(=\)	\(2>0\)

\(f''>0\), so \(x=2\) is a local minimum; the value is \(f(2)=-3\).

local minimum at x equals 2, value minus 3

Example 2 — A cubic with both

Find and classify the extrema of \(f(x)=x^3-3x\).

Solution

Critical points from \(f'=3x^2-3=0\) are \(x=\pm 1\); test with \(f''=6x\).

\(f''(1)\)	\(=\)	\(6>0\ \Rightarrow\ \text{min}\)
\(f''(-1)\)	\(=\)	\(-6<0\ \Rightarrow\ \text{max}\)

Local max at \(x=-1\), local min at \(x=1\).

local max at minus one, local min at one

Example 3 — First-derivative test

Classify the critical point of \(f(x)=x^3\) at \(x=0\).

Solution

\(f'(x)=3x^2\ge 0\) on both sides of \(0\) — no sign change.

\(f'(-1)\)	\(=\)	\(3>0\)
\(f'(1)\)	\(=\)	\(3>0\)

No sign change, so \(x=0\) is neither a max nor a min.

neither maximum nor minimum at x equals 0

Example 4 — The maximum value

Find the local maximum value of \(f(x)=-x^2+6x-5\).

Solution

Critical point from \(f'=-2x+6=0\) is \(x=3\); \(f''=-2<0\) confirms a max.

\(f(3)\)

\(=\)

\(-9+18-5=4\)

The local maximum value is \(4\).

local maximum value is 4

Common pitfalls

\(f''(c)=0\) is inconclusive. The second-derivative test fails there; use the first-derivative test instead.

A critical point can be neither. \(x^3\) has \(f'(0)=0\) but no extremum — \(f'\) never changes sign.

Report the value, not just the location. The maximum/minimum value is \(f(c)\), which needs a substitution.

Frequently asked questions

How do you find local maxima and minima?

Solve \(f'(x)=0\) for critical points, then classify each with the second-derivative test or a first-derivative sign chart.

What is the second-derivative test?

At a critical point \(c\): if \(f''(c)>0\) the point is a local minimum; if \(f''(c)<0\) it is a local maximum. If \(f''(c)=0\) the test is inconclusive.

What is the first-derivative test?

Check how \(f'\) changes sign at the critical point: \(+\) to \(-\) is a max, \(-\) to \(+\) is a min, no change is neither.

What is the difference between the location and the value of a maximum?

The location is the \(x\)-value \(c\) where it occurs; the value is \(f(c)\), the height of the curve there.

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