Use a graph or table to identify a guarantee
Four readings of a river
A table gives the depth of a river, in meters, at four times. The dots show the readings: 2.0 at hour 0, 3.1 at hour 2, 4.9 at hour 4 and 5.6 at hour 6.
The problem states that the depth is continuous and differentiable. The curve is one possible graph through the dots.
The solid line joins the readings at hours 0 and 6 and is called the secant line. It rises 3.6 m in 6 h, so its slope is 0.6 m/h.
The orange point carries its tangent and a dashed copy of the secant line. Drag the point until the tangent lies along the dashed line. Then change the shape of the curve. The dots and the secant line stay, but the parallel tangent moves to a new hour.
Readings and stated conditions
The numbers are gone. The two end readings, \(f(a)\) and \(f(b)\), fix the secant line, and a reading between them does not change it.
The table does not show the curve between the readings, and a differentiable graph can bend in many ways. Move the slider to see some.
If the problem states that \(f\) is continuous and differentiable, then for every shape some tangent is parallel to the secant line. Move the point with the slider for x to find it.
Notation
Read the two values at the ends of the interval from the table or the graph.
\[f(a), \quad f(b)\]Divide the change in \(f\) by the change in \(x\). This is the average rate of change, and its units are the units of \(f\) per unit of \(x\).
\[\cy{m} = \dfrac{f(b) - f(a)}{b - a}\]Find the conditions in the statement of the problem. It must say that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b)\).
Readings cannot show these conditions. The words "f is differentiable for all x" give both, because a differentiable function is continuous.
If both conditions are stated, the theorem guarantees some \(c\) in \((a, b)\) with this derivative. The table does not say which \(c\) it is.
\[f'(c) = \cy{m}\]
When it fails
The same four readings also fit three straight pieces with corners at hours 2 and 4, where there is no tangent. Every other tangent lies along a piece with slope 0.55, 0.9 or 0.35 m/h, so none has slope 0.6 m/h.
A table cannot show that a curve is smooth, so the problem must state the conditions. A jump between two readings would break the guarantee in the same way.
Example 1: a table of h
A table gives values of \(h\). The problem states that \(h\) is continuous on \([0, 9]\) and differentiable on \((0, 9)\). Does the theorem guarantee a point with \(h'(c) = 2\)?\[\begin{array}{c|cccc} x & 0 & 2 & 5 & 9 \\ \hline h(x) & 3 & 5 & 6 & 21 \end{array}\]
Both conditions are stated, so the theorem applies.
The average rate of change of \(h\) over \([0, 9]\) is 2.
\[\cy{m} = \dfrac{h(9) - h(0)}{9 - 0} = \dfrac{21 - 3}{9} = 2\]Yes. Some \(c\) in \((0, 9)\) has \(h'(c) = 2\).
The curve is only one possible graph, so it shows one possible place for c.
Example 2: readings of a cart
A table gives the position \(p\) of a cart, in meters, at time \(t\) seconds. The problem states that \(p\) is continuous on \([0, 8]\) and differentiable on \((0, 8)\). Is there a time when the speed is 1.5 m/s?\[\begin{array}{c|ccccc} t & 0 & 2 & 4 & 6 & 8 \\ \hline p(t) & 0 & 1 & 2 & 3 & 12 \end{array}\]
Both conditions are stated, so the theorem applies.
A tempting wrong move: The rates between neighboring readings are 0.5, 0.5, 0.5 and 4.5 m/s. None of them is 1.5 m/s, so there is no guarantee.
\[\begin{aligned} \dfrac{1 - 0}{2 - 0} &= 0.5 \\ \dfrac{2 - 1}{4 - 2} &= 0.5 \\ \dfrac{3 - 2}{6 - 4} &= 0.5 \\ \dfrac{12 - 3}{8 - 6} &= 4.5 \end{aligned}\]These four lines join neighboring readings, not the two ends of the interval. The theorem uses the one secant line from \((0, 0)\) to \((8, 12)\), and its slope is 1.5.
Use the two end readings. The average speed is 1.5 m/s, so at some time \(c\) in \((0, 8)\) the speed \(p'(c)\) is 1.5 m/s.
\[\cy{m} = \dfrac{12 - 0}{8 - 0} = \cy{1.5}\]
Practice
A table gives values of \(k\). The problem states that \(k\) is continuous on \([1, 10]\) and differentiable on \((1, 10)\). On which interval does the theorem guarantee a point with \(k'(c) = 2\): \((1, 3)\), \((3, 6)\), \((6, 10)\) or \((3, 10)\)?\[\begin{array}{c|cccc} x & 1 & 3 & 6 & 10 \\ \hline k(x) & 4 & 10 & 13 & 24 \end{array}\]
For each interval, divide the change in \(k\) by the change in \(x\). Compare the result with 2.
Solution
The conditions stated for [1, 10] are also true on each smaller interval, so the theorem applies to all four. Find the average rate of change on each.
\[\begin{aligned} \dfrac{10 - 4}{3 - 1} &= 3, & \dfrac{13 - 10}{6 - 3} &= 1 \\ \dfrac{24 - 13}{10 - 6} &= \dfrac{11}{4}, & \dfrac{24 - 10}{10 - 3} &= 2 \end{aligned}\]Only \((3, 10)\) has the average rate 2, so the theorem guarantees \(k'(c) = 2\) there.
Questions
The graph shows \(g\), continuous on \([0, 6]\) and differentiable on \((0, 6)\), with \(g(0) = 2\), \(g(2) = 3\), \(g(4) = 7\) and \(g(6) = 8\).
What slope does the theorem guarantee for some tangent between \(x = 0\) and \(x = 6\)?
- 1
- 6
- \(\tfrac43\)
Answer
1
On which interval does the theorem guarantee a point with \(g'(c) = 2\)?
- \((2, 4)\)
- \((0, 4)\)
- \((0, 2)\)
- \((4, 6)\)
Answer
\((2, 4)\)
A table gives values of \(r\), and \(r\) is differentiable for all \(x\). Which value of \(r'\) does the theorem guarantee at some point of \((4, 10)\)?\[\begin{array}{c|ccc} x & 0 & 4 & 10 \\ \hline r(x) & 5 & 17 & 29 \end{array}\]
- 2
- 3
- 12
- 2.4
Answer
2
Summary
You read the two end values, find the average rate of change, and confirm that the problem states the conditions.
This works because the end readings fix the secant line, and the conditions guarantee a parallel tangent.
You use it when a problem gives a table or a graph and asks about a rate that the function must reach.
This is one of five skills in topic 5.1 of Which & Why: AP Calculus.