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Lecture 16: Relative Extrema and Critical Numbers

Goal: Discuss when a function's behavior changes.

Two-way diagram between f and f prime labeled with velocity, growth rate, related rates and extrema, concavity.
Figure 1
A diagram showing two curved arrows forming a loop between the symbol f on the left and f prime on the right. The upper arrow points from f to f prime and is labeled Velocity, Growth Rate, Related Rates. The lower arrow points back from f prime to f and is labeled below with Extrema, Intervals of Inc./Dec., Concavity, Inflection Points.

When answering the question "why do we take derivatives?" there are two main answers.

  1. If we are given a function \(f\), \(f'\) represents how that function changes. We saw this when we talked about velocity, acceleration, growth rates, and related rates.
  2. The reverse is also true: if we know how a function changes (represented by \(f'\)), then we can recover information about our original function \(f\).

We want to start thinking about Bullet Point 2. To do so, consider the following graph:

Hand-drawn curve on axes with labeled local maxima and local minima and an interval I around c.
Figure 2
A hand-drawn graph with a horizontal x-axis and a vertical y-axis. A wavy curve enters from the upper left, descends to a point circled in blue labeled Local Min, rises to a peak circled in red labeled Local Max, then descends crossing the y-axis through a red circled point, continues down to a point circled in blue just above the x-axis labeled Local Min, rises to another red circled peak labeled Local Max, then descends to a blue circled point sitting on the x-axis labeled Local Min, and finally rises steeply to the upper right. Beneath the first local maximum a red bracketed interval is drawn on the x-axis with the label c on the left and I on the right, indicating the point c inside the interval I.

In Figure 1, there is a graph and 6 "points of interest." I.e., there are 6 locations where the behavior of the function changes. We want to talk them, so we give them special names.

Definition: A function \( f \) has a relative maximum (or local maximum) at \( x=c \) if there is an interval \( I \) containing \( c \) such that \( f(x) \le f(c) \) for \( x \) in \( I \).

Definition: A function \( f \) has a relative minimum (or local minimum) at \( x=c \) if there is an interval \( I \) containing \( c \) such that \( f(x) \geq f(c) \) for \( x \) in \( I \).

Definition: A point is a relative extrema if it is a relative maximum or minimum.

The relative extrema on Figure 2 are labeled accordingly. Ideally, we want to classify all local extrema of a function given its formula (not just its graph). We have a theorem to deal with this:

Theorem (Fermat, \( \sim 1636 \)): If \( f \) has a relative maximum or minimum at a point \( c \), then \(f'(c)\) is either \(0\) or undefined.

NOTE:

The reverse is not true. Take \(f(x) = x^3\) and \(g(x) = \sqrt[3]{x}\), \(f'(0) = 0\) and \(g'(0)\) is undefined, but taking a look at the graphs (in Figures 3 and 4) neither function has a relative maximum or minimum at \(0\).

Sketch of the graph of x cubed with a marked point at the origin where the curve flattens.
Figure 3
A small blue hand-drawn sketch of coordinate axes with the graph of \( x^3 \), labeled x cubed to the right of the curve. The curve rises from lower left to upper right, flattening as it passes through the origin, where a point is circled to indicate that the derivative is zero there even though the point is neither a maximum nor a minimum.
Sketch of the graph of x to the one third power with a marked point at the origin with vertical tangent.
Figure 4
A small blue hand-drawn sketch of coordinate axes with the graph of \( x^{1/3} \), labeled to the upper right. The curve rises from lower left to upper right, passing through the origin with a vertical tangent, and the origin is circled to indicate that the derivative is undefined there even though the point is neither a maximum nor a minimum.

Unfortunately, we can not determine the locations of relative extrema using Fermat's Theorem alone. However, it lets us pinpoint the locations of them to certain points, then we can do more ivestigative work. The locations where \(f'(x)=0\) or undefined are important enough we give them their own name.

Definition: Let \( c \) be in the domain of \( f \). If \( f'(c) \) is \( 0 \) or undefined, we say \( c \) is a critical number (or critical point or critical value) of \( f \).

Example 1

Let \( f(x) = 2\cos(5x) + 5x \)

  1. Is \( x = \frac{\pi}{30} \) a critical number for \( f \)?

    \[ \begin{gathered} f'(x) = -10\sin(5x) + 5 \\[0.5em] f'\left(\frac{\pi}{30}\right) = -10\sin\left(5 \cdot \frac{\pi}{30}\right) + 5 = -10\sin\left(\frac{\pi}{6}\right) + 5 \\[0.5em] = -10\left(\frac{1}{2}\right) + 5 = 0 \ \checkmark \ \text{yes} \end{gathered} \]
    Right triangle with angle pi over 6, hypotenuse 2, opposite side 1, adjacent side square root of 3.
    Figure 5
    A hand-drawn reference right triangle in red. The hypotenuse is labeled 2 along the upper-left side, the angle at the left vertex is labeled \( \pi/6 \), the vertical side on the right is labeled 1, and the horizontal base is labeled \( \sqrt{3} \). It is used to evaluate \( \sin(\pi/6) = \frac{1}{2} \).

    2. So \(\frac{\pi}{30}\) IS a critical number by definition.

  2. Is \( x = \pi \) a crit. num. for \( f \)?

    \[ \begin{gathered} f'(x) = -10\sin(5x) + 5 \\[0.5em] f'(\pi) = -10\sin(5\pi) + 5 \\[0.5em] = 5 \ne 0 \quad \text{or undefined} \end{gathered} \]

    So \(\pi\) is NOT a critical number.

Example 2

Find all critical numbers for

\[ \begin{gathered} f(x) = x^3 - 75x\end{gathered} \]

\( f'(x) = 3x^2 - 75 \) is never undefined. So, we need to solve

\[ \begin{gathered} 3x^2 - 75 \overset{\text{set}}{=} 0 \\[0.5em] 3(x^2 - 25) = 0 \\[0.5em] 3(x-5)(x+5) = 0 \\[0.5em] \Rightarrow x = -5, 5 \end{gathered} \]

Example 3

Find all critical numbers for

\[ \begin{gathered} f(x) = x^3 + 5x^2 + 7x \end{gathered} \]

\( f'(x) = 3x^2 + 10x + 7 \) is never undefined. So, we do the following:

\[ \begin{gathered} {3}x^2 + {10}x + {7} \overset{\text{set}}{=} 0 \\[0.5em] x = \frac{-10 \pm \sqrt{(10)^2 - 4(3)(7)}}{2(3)} = \frac{-10 \pm \sqrt{100 - 84}}{6} \\[0.5em] = \frac{-10 \pm \sqrt{16}}{6} = \frac{-10 \pm 4}{6} \Rightarrow x = \frac{-10 + 4}{6}, \ x = \frac{-10 - 4}{6} \\[0.5em] \Rightarrow x = -1, -\frac{7}{3} \end{gathered} \]

NOTE:

Here we used the quadratic formula. Here it is for reference, \[ ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Example 4

Find all critical numbers of

\[ \begin{gathered} f(x) = 4x^3 - \frac{4}{x^3} \\[0.5em] f'(x) = 12x^2 - \left[4x^{-3}\right]' = 12x^2 + 12x^{-4} \\[0.5em] = 12x^2 + \frac{12}{x^4} \end{gathered} \]

NOTE:

It is tempting to say \( 0 \) is a critical number here since \( f'(0) \) is undefined. But remember, critical numbers are in the domain of the function, which \( 0 \) is not.

\( f' \) is never undefined (in the domain of \( f \)).

So, we solve \( f'(x) = 0 \)

\[ \begin{gathered} 12x^2 + \frac{12}{x^4} \overset{\text{set}}{=} 0 \\[0.5em] x^4\left(12x^2 + \frac{12}{x^4}\right) = 0 \cdot x^4 \\[0.5em] 12x^6 + 12 = 0 \\[0.5em] x^6 + 1 = 0 \end{gathered} \]

\(x^6+1=0\) has NO Real Solutions. Hence, \(f\) has no critical numbers.

Sketch of U-shaped curve for x to the sixth plus 1 above dashed line y equals 1.
Figure 7
A freehand sketch of a steep U-shaped curve, labeled x to the sixth plus 1, drawn above a horizontal dashed line labeled y = 1. A small circled point marks where the bottom of the U touches the dashed line, illustrating that the curve never reaches zero and so x to the sixth plus 1 equals 0 has no real solution.

Example 5

Repeat for

\[ \begin{gathered} f(x) = xe^x \end{gathered} \]

\( f'(x) = (1)e^x + xe^x \) is never undefined. So, we solve

\[ \begin{gathered} 1 \cdot e^x + xe^x \overset{\text{set}}{=} 0 \\[0.5em] e^x(1+x) = 0 \\[0.5em] e^x = 0 \quad \textbf{OR} \quad 1+x = 0 \end{gathered} \]

As \(e^x\) is never 0, we only need to solve \(1 + x = 0\). Hence, \(-1\) is the only critical number.

Example 6

Repeat for \(f(x) = 6x^4 e^{5x+4}\).

\[ \begin{gathered} f'(x) = 24x^3 e^{5x+4} + 6x^4\left[5 \cdot e^{5x+4}\right] \\[0.5em] = 24x^3 e^{5x+4} + 30x^4 e^{5x+4} \overset{\text{set}}{=} 0 \\[0.5em] 6x^3 e^{5x+4}\left[4 + 5x\right] = 0 \\[0.5em] 6x^3 = 0 \quad \textbf{OR} \quad e^{5x+4} = 0 \quad \textbf{OR} \quad 4 + 5x = 0 \end{gathered} \]

\(e^{5x+4}\) is never \(0\). So, solving \(6x^3 = 0\) we get \(x = 0\) and solving \(4 + 5x = 0\) we get \(x = -\frac{4}{5}\). Hence, the only critical numbers are \(0\) and \(-\frac{4}{5}\).

Example 7

Repeat for \( f(x) = \frac{\ln x}{x} \)

\[ f'(x) = \frac{\frac{1}{x} \cdot x - (1)\ln x}{x^2} = \frac{1 - \ln x}{x^2} \]

\( f'(x) \) is never undefined (in the domain of \( f \))

\[ \begin{gathered} \frac{1 - \ln x}{x^2} = 0 \\[0.5em] 1 - \ln x = 0 \\[0.5em] \ln x = 1 \\[0.5em] \boxed{x = e} \end{gathered} \]
Sketch of y equals ln x over x with a peak at x equals e marked by a dotted line.
Figure 9
A hand-drawn sketch of the graph of \( f(x) = \frac{\ln x}{x} \) on horizontal and vertical axes. The curve comes up steeply from far below near the vertical axis on the right side of the origin, crosses the horizontal axis just to the right of the vertical axis, rises to a single peak, and then decreases gradually, flattening out toward the horizontal axis as x increases. A vertical dotted line drops from the peak down to the horizontal axis, where the point is labeled e, indicating the maximum occurs at \( x = e \).