AP Calculus AB/BC

Part 5 · analytical-applications

Extrema and the shape of a graph

The first derivative finds the turning points, the second decides which kind they are, and together they draw the curve.

By the end of this chapter you can

  • Find critical points and classify them
  • Use the first and second derivative tests, and know when each fails
  • Locate inflection points and relate them to concavity
  • Apply the Extreme Value Theorem on a closed interval

A derivative is a slope, so it knows where a curve is rising, falling, and momentarily doing neither. That last case is where maxima and minima live, and extracting the full shape of a graph from f′f' and f′′f'' is one of the most heavily tested skills in AB.

Critical points

cc is a critical point when f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist.

Those are the only places an interior extremum can occur — which is worth stating precisely, because the converse is false and the exam tests the difference.

  • ddx(x3−3x)=3x2−3\frac{d}{d x} \left(x^{3} - 3 x\right) = 3 x^{2} - 3
  • ddxx3=3x2\frac{d}{d x} x^{3} = 3 x^{2}
checked by the build

For x3−3xx^3 - 3x, setting 3x2−3=03x^2 - 3 = 0 gives x=±1x = \pm 1: two critical points, and both turn out to be extrema.

For x3x^3, setting 3x2=03x^2 = 0 gives x=0x = 0 — a critical point that is not an extremum. The curve flattens momentarily and carries straight on up.

The first derivative test

Look at the sign of f′f' on either side of the critical point:

f′f' goes at cc there is
++ then −- a local maximum
−- then ++ a local minimum
no sign change neither

This is the test that always works, and it is the one to fall back on. It explains x3x^3 immediately: f′=3x2f' = 3x^2 is positive on both sides of zero, no sign change, so no extremum.

The second derivative test

Faster when it applies:

f′(c)=0 and f′′(c)>0  ⇒  minimumf'(c) = 0 \text{ and } f''(c) > 0 \;\Rightarrow\; \text{minimum}

f′(c)=0 and f′′(c)<0  ⇒  maximumf'(c) = 0 \text{ and } f''(c) < 0 \;\Rightarrow\; \text{maximum}

  • ddx(3x2−3)=6x\frac{d}{d x} \left(3 x^{2} - 3\right) = 6 x
checked by the build

For x3−3xx^3 - 3x, f′′=6xf'' = 6x. At x=1x = 1 it is +6+6: a minimum. At x=−1x = -1 it is −6-6: a maximum. Two substitutions and the classification is done.

  • ddxx4=4x3\frac{d}{d x} x^{4} = 4 x^{3}
  • ddx4x3=12x2\frac{d}{d x} 4 x^{3} = 12 x^{2}
checked by the build

At x=0x = 0 that second derivative is 00 — and x4x^4 plainly has a minimum there. Verified, and a reminder that the fast test is not the reliable one.

Concavity and inflection

f′′f'' describes how the slope itself is changing:

  • f′′>0f'' > 0 — slope increasing — concave up, curving like a cup.
  • f′′<0f'' < 0 — slope decreasing — concave down, like a cap.

An inflection point is where concavity changes sign. As with extrema, f′′=0f'' = 0 is necessary but not sufficient: x4x^4 has f′′(0)=0f''(0) = 0 and no inflection there, because f′′=12x2f'' = 12x^2 never changes sign.

The pattern repeats one level up, and noticing that is worth more than memorising both cases: a zero of a derivative is a candidate, and a sign change is the evidence.

Drag hh towards zero with the pivot at x=−1x = -1: the secant slope heads for zero, which is what a turning point looks like from the derivative’s side.

The Extreme Value Theorem

If ff is continuous on a closed interval [a,b][a,b], it attains an absolute maximum and an absolute minimum somewhere on it.

That guarantee makes the closed-interval method work:

  1. Find the critical points inside (a,b)(a, b).
  2. Evaluate ff at those, and at both endpoints.
  3. The largest value is the absolute maximum; the smallest is the minimum.

No derivative test is needed — comparing the numbers settles it.