Lesson 16: Relative Extrema and Critical Numbers
We begin by formally defining what it means for a function to have a relative (local) maximum or minimum at a point.
Definition (Relative Maximum): If \(f(c) \geq f(x)\) for all \(x\) in an open interval \(I\) containing \(c\), then \(f(c)\) is a relative maximum of \(f\).
Definition (Relative Minimum): If \(f(c) \leq f(x)\) for all \(x\) in an open interval \(I\) containing \(c\), then \(f(c)\) is a relative minimum of \(f\).
Together, relative maxima and relative minima are called relative extrema.
Graph Description: A continuous curve is plotted on an \(xy\)-plane illustrating five points of interest. The point \((-7, -2)\) is annotated with \(f'(-7) = 0\), but it is not a relative extremum, illustrating the important caveat that a zero derivative does not guarantee an extremum. At \((-3, 6)\), the curve reaches a local peak (relative maximum) where \(f'(-3) = 0\). At \((1, -5)\), the curve reaches a local valley (relative minimum) where \(f'(1) = 0\). At \((3, 3)\), there is another relative maximum, but here \(f'(3)\) is undefined — the graph has a sharp corner or cusp at this point. At \((6, -1)\), the curve does not have a relative extremum, and \(f'(6)\) is also undefined.
Warning: A relative maximum or minimum must occur at a point where the derivative is zero or does not exist. However, a point where \(f'(x) = 0\) or \(f'(x)\) does not exist is not necessarily a relative maximum or minimum. Both conditions must be checked carefully.
Critical Numbers
Definition (Critical Number): Let \(c\) be a number in the domain of the function \(f(x)\). If \(f'(c) = 0\) or \(f'(c)\) does not exist, then \(c\) is a critical number of \(f(x)\).
Key Idea: To find the relative extrema of a function, we first find all critical numbers, then test those critical numbers to determine whether each gives a relative maximum, relative minimum, or neither. The testing procedures will be covered in upcoming lessons.
Examples
Example 1
Find the critical numbers of \(y = x^3 + \dfrac{3}{2}x^2 - 36x\).
Solution:
We begin by differentiating:
\[y' = 3x^2 + 3x - 36\]To find critical numbers, we solve \(y' = 0\) and check where \(y'\) does not exist (DNE). Note that since \(y'\) is a polynomial, it is defined everywhere. So, we just need to find where \(y'=0\).
\[0 = 3x^2 + 3x - 36= 3(x^2 + x - 12)\implies 0 = (x + 4)(x - 3)\implies x = -4,3\]The critical numbers are \(x = -4\) and \(x = 3\).
Example 2
Find the critical numbers of \(y = 3x^2 - \dfrac{4}{x^2}\).
Solution:
Rewriting \(y = 3x^2 - 4x^{-2}\) and differentiating:
\[y' = 6x + 8x^{-3} = 6x + \frac{8}{x^3}\]Note that \(y'\) is undefined when \(x = 0\). However, \(x = 0\) is not in the domain of the original function \(y\) (since \(\frac{4}{x^2}\) is undefined at \(x = 0\)). Therefore, \(x = 0\) is not a critical number.
Now, let's solve \(y'=0\):
\[0 = 6x + \frac{8}{x^3}\implies -6x = \frac{8}{x^3}\implies x^4 = -\frac{8}{6} = -\frac{4}{3}\]Since \(x^4 \geq 0\) for all real \(x\), but the right-hand side is negative, this equation has no real solutions. Therefore, \(y = 3x^2 - \dfrac{4}{x^2}\) has no critical numbers.
Example 3
Find the critical numbers of \(y = 2x^4 e^{3x}\).
Solution:
Using the product rule:
\[y' = 8x^3 \cdot e^{3x} + 3e^{3x} \cdot 2x^4 = 8x^3 e^{3x} + 6x^4 e^{3x}\]Note that \(y'\) is defined for all real \(x\). Solving \(y' = 0\):
\[0 = 8x^3 e^{3x} + 6x^4 e^{3x} = 2x^3 e^{3x}(4 + 3x)\]So, either \(2x^3 = 0\), \(e^{3x} = 0\), or \(4 + 3x = 0\). If \(2x^3 = 0\), we have that \(x = 0\). Notice that \(e^{3x} = 0\) is impossible, since \(e^{3x} > 0\) for all \(x\). If \(4 + 3x = 0\), then \(x = -\dfrac{4}{3}\). So, the critical numbers are \(x = 0\) and \(x = -\dfrac{4}{3}\).
Example 4
Find the critical numbers of \(y = \dfrac{7x^2 + 12}{3x}\).
Solution:
Using the quotient rule:
\[y' = \frac{14x \cdot 3x - 3(7x^2 + 12)}{(3x)^2} = \frac{42x^2 - 21x^2 - 36}{9x^2} = \frac{21x^2 - 36}{9x^2}\]Notice that \(y'\) is undefined when \(x = 0\). However, \(x = 0\) is not in the domain of the original function \(y\) (since the denominator cannot be zero). Therefore, \(x = 0\) is not a critical number. Let's solve for \(y'=0\).
\[0 = \frac{21x^2 - 36}{9x^2} \implies 0 = 21x^2 - 36 \implies x^2 = \frac{36}{21}\]Continuing,
\[x = \pm\sqrt{\frac{36}{21}} = \pm\frac{6}{\sqrt{21}} = \pm\frac{6\sqrt{21}}{21} = \pm\frac{2\sqrt{21}}{7} = \pm\frac{2\sqrt{3}}{\sqrt{7}}.\]The critical numbers are \(x = \dfrac{2\sqrt{3}}{\sqrt{7}}\) and \(x = -\dfrac{2\sqrt{3}}{\sqrt{7}}\).
Example 5
Find the critical numbers of \(y = 2\cos(4x) + 4x\) on the interval \((0, \pi)\).
Solution:
Differentiating:
\[y' = -8\sin(4x) + 4\]Note that \(y'\) is defined everywhere. Solving \(y' = 0\):
\[0 = -8\sin(4x) + 4 \implies \sin(4x) = \frac{1}{2}\]On the unit circle, \(\sin(\theta) = \frac{1}{2}\) when \(\theta = \frac{\pi}{6}\) or \(\theta = \frac{5\pi}{6}\). So,
\[4x = \frac{\pi}{6} \quad \Rightarrow \quad x = \frac{\pi}{24}\] \[4x = \frac{5\pi}{6} \quad \Rightarrow \quad x = \frac{5\pi}{24}\]Both values lie in \((0, \pi)\), so the critical numbers are \(x = \dfrac{\pi}{24}\) and \(x = \dfrac{5\pi}{24}\).
Unit Circle Reference: The unit circle diagram shows the angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\). At angles \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\), the sine value equals \(\frac{1}{2}\). The cosine values are \(\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\), respectively. These are the two standard angles in \([0, \pi]\) where \(\sin(\theta) = \frac{1}{2}\).