FUNCTIONS · GRAPH ANALYSIS

Extrema & graph analysis

Maxima and minima describe where a function is largest or smallest. Derivatives turn this visual idea into a systematic method for locating and classifying turning behaviour.

local maximumlocal minimum

1. Local versus global extrema

A point \(x=c\) is a local maximum if \(f(c)\) is at least as large as nearby function values. It is a local minimum if it is at most as large as nearby values. A global extremum compares \(f(c)\) with every value in the full domain under consideration.

Local maximumHighest value only in a neighbourhood of \(c\).
Local minimumLowest value only in a neighbourhood of \(c\).
Global extremumHighest or lowest value on the entire chosen domain.

2. Critical points

If \(f\) has a local extremum at an interior point \(c\) and is differentiable there, then Fermat's theorem gives

\[f'(c)=0\]

So candidates for extrema occur where \(f'(x)=0\), or where the derivative does not exist while the function itself does. These are critical points.

Important: \(f'(c)=0\) is necessary for a differentiable interior extremum, but not sufficient. For example, \(f(x)=x^3\) has \(f'(0)=0\) but no maximum or minimum at \(0\).

3. Classifying critical points

First-derivative test

Inspect the sign of \(f'(x)\) immediately to the left and right of a critical point.

+ → −Increasing then decreasing: local maximum.
− → +Decreasing then increasing: local minimum.
No sign changeUsually not an extremum.

Second-derivative test

If \(f'(c)=0\) and \(f''(c) eq0\), curvature gives a faster classification:

\[f''(c)>0\Rightarrow\text{local minimum},\qquad f''(c)<0\Rightarrow\text{local maximum}.\]

If \(f''(c)=0\), the test is inconclusive; use the first-derivative test or another argument.

4. Finding global extrema on a closed interval

For a continuous function on \([a,b]\), the Extreme Value Theorem guarantees an absolute maximum and minimum. To find them:

1
Find all critical points inside \((a,b)\).
2
Evaluate \(f\) at every critical point.
3
Evaluate the endpoints \(f(a)\) and \(f(b)\).
4
Compare all resulting values. Largest = global maximum, smallest = global minimum.

5. Worked example

Analyze \(f(x)=x^3-3x\).

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

The critical points are \(x=-1\) and \(x=1\). Since

\[f''(x)=6x,\]

we get \(f''(-1)=-6<0\), so \(x=-1\) is a local maximum, and \(f''(1)=6>0\), so \(x=1\) is a local minimum.

\[f(-1)=2,\qquad f(1)=-2.\]

Therefore the local maximum is \((-1,2)\) and the local minimum is \((1,-2)\).

6. Interactive cubic extrema analyzer

Enter \(f(x)=ax^3+bx^2+cx+d\). The analyzer solves \(f'(x)=0\) and classifies real critical points using \(f''(x)\).

Example

\(x^3-3x\) has a local maximum at \(x=-1\) and a local minimum at \(x=1\).

7. Common mistakes

Solving only \(f'(x)=0\). Critical points are candidates; they still need classification.
Forgetting endpoints. On a closed interval, global extrema can occur at endpoints even when \(f'(x) eq0\).
Treating \(f''(c)=0\) as proof of an inflection point. It is only a candidate; concavity must actually change.
Confusing x-values and function values. An extremum point is \((c,f(c))\), not just \(c\).

8. Practice

  1. Classify the critical points of \(f(x)=x^4-4x^2\).
  2. Find the global maximum and minimum of \(f(x)=x^2-4x+1\) on \([0,5]\).
  3. Explain why \(f(x)=x^3\) has a stationary point at \(0\) but no extremum there.
Solutions

1) \(f'(x)=4x(x^2-2)\): local maximum at \(x=0\), local minima at \(x=\pm\sqrt2\). 2) Vertex at \(x=2\): minimum \(f(2)=-3\); endpoints give \(f(0)=1\), \(f(5)=6\), so maximum is \(6\) at \(x=5\). 3) \(f'(0)=0\), but \(f'(x)>0\) on both sides, so the function keeps increasing.