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Maxima and minima

Critical points and the second-derivative test.

A smooth function peaks or bottoms out only where its derivative is zero. Find those critical points, then use the second derivative to tell a maximum (f″ < 0) from a minimum (f″ > 0). This is the whole of optimisation in one variable.

Worked example · critical points of x^3 - 3x

Critical points of x^3 - 3x

x^{3} - 3 x

Step by step

  1. f(x) = x^{3} - 3 x

    Critical points are where f′(x) = 0 or is undefined.

  2. f'(x) = 3 x^{2} - 3

    Differentiate.

  3. x = -1, x = 1

    Solve f′(x) = 0.

  4. f''(x) = 6 x

    Second-derivative test: f″ < 0 → maximum, f″ > 0 → minimum.

  5. f''(-1) = -6 \Rightarrow (-1, 2) \text{ is a local maximum}

  6. f''(1) = 6 \Rightarrow (1, -2) \text{ is a local minimum}

Reveal the answer
(-1, 2)\ \text{local maximum},\; (1, -2)\ \text{local minimum}

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How to: Maxima and minima

  1. Critical points are where f′(x) = 0 or is undefined.
  2. Differentiate.
  3. Solve f′(x) = 0.
  4. Second-derivative test: f″ < 0 → maximum, f″ > 0 → minimum.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point, the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate: the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function, a polynomial times an exponential, log or trig function.

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