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Critical Point

Critical Point – definition and example, AP Calculus AB math glossary – Mathovore

Key takeaway: A critical point of a function is an input in its domain where the derivative equals zero or does not exist, which makes it a candidate for a local maximum or minimum.

Grade 12 – AP Calculus ABAP Calculus AB Unit 5Calculus

Definition of Critical Point

Definition

A critical point marks a place where a graph might turn around. At a smooth peak or valley the tangent line is horizontal, so the derivative is zero. At a sharp corner or cusp the derivative does not exist. Both kinds of locations are flagged for further study.

The input must belong to the domain of the function itself. Being critical does not guarantee an extreme value: some critical points are only flat stretches in the middle of a rise or fall.

Formula and notation

The number \(k\) is a critical number of \(v\) when \(v(k)\) exists and

\[ v^{\prime}(k)=0\quad\text{or}\quad v^{\prime}(k)\text{ is undefined}. \]

The matching point on the graph is \((k,\,v(k))\). To find them, set the derivative equal to zero and also check where its denominator vanishes.

x = 0: cusp, f’ undefinedx = 2: f’ = 0x
Two critical points of one function: a cusp where the derivative fails to exist and a valley where it equals zero.

Key facts

  • Every local maximum or minimum of a differentiable function sits at a critical point.
  • Not every critical point is an extreme: y = x cubed has one at 0 but no peak or valley.
  • Where the derivative is undefined counts only if the function itself is defined there.
  • These inputs divide the number line into intervals for a sign check of the derivative.

Critical Point example

Problem

Find all critical points of \(f(x)=x^{5/3}-5x^{2/3}\).

  1. Differentiate: \(f^{\prime}(x)=\frac{5}{3}x^{2/3}-\frac{10}{3}x^{-1/3}=\frac{5(x-2)}{3x^{1/3}}\).

  2. The derivative equals zero when the numerator does: \(x=2\).

  3. It is undefined when the denominator is zero: \(x=0\). Since \(f(0)=0\) exists, \(x=0\) qualifies.

The function has two critical points, at \(x=0\) and \(x=2\), with outputs \(f(0)=0\) and \(f(2)\approx -4.76\).

Common mistake

Watch out

A student solves only \(f^{\prime}(x)=0\) for the function above and reports a single critical point at \(x=2\).

Why it is wrong

That misses the cusp at \(x=0\). They come from two sources: zeros of the derivative and places where the derivative is undefined but the function is defined. Always examine the denominator of the derivative as well as its numerator.

Related terms and lessons

Browse the glossary: A–Z list · AP Calculus AB vocabulary · Calculus

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