Following Secant Slopes on Shorter Intervals
A secant slope measures the average change in a function's output across an interval of inputs. If we keep one input fixed and move the other closer to it, the interval shrinks. Recording the corresponding slopes in a table can show whether the values appear to settle near a particular number.
This numerical investigation is useful, but it has a clear boundary: a finite table suggests a pattern; it does not establish that the slopes approach an exact value. Here we focus on making and interpreting the table carefully, without treating its evidence as a proof of what happens for every smaller interval.
When \(h\) is positive, the second input \(a+h\) is greater than \(a\). If we choose a sequence of positive \(h\)-values that get smaller, we can compare the slopes over progressively shorter intervals starting at \(a\). Each table row needs the interval width, the two endpoint outputs, the output change, and the resulting quotient. Keep enough digits to see the trend before rounding for display.
Worked Example: A Negative Slope Near \(-0.25\)
Worked Example: A Negative Slope Near \(-0.25\)
Take \(f(x)=1/x\), fix \(a=2\), and calculate slopes using \(h=0.5,\ 0.2,\ 0.1,\ 0.05\). At the fixed input, \(f(2)=1/2=0.5\). For each row, the other endpoint is \(2+h\).
| \(h\) | \(2+h\) | \(f(2+h)\) | Output change | Secant slope |
|---|---|---|---|---|
| 0.5 | 2.5 | 0.4 | 0.4 − 0.5 = −0.1 | −0.1 ÷ 0.5 = −0.200000 |
| 0.2 | 2.2 | 0.454545 | 0.454545 − 0.5 = −0.045455 | −0.045455 ÷ 0.2 ≈ −0.227273 |
| 0.1 | 2.1 | 0.476190 | 0.476190 − 0.5 = −0.023810 | −0.023810 ÷ 0.1 ≈ −0.238095 |
| 0.05 | 2.05 | 0.487805 | 0.487805 − 0.5 = −0.012195 | −0.012195 ÷ 0.05 ≈ −0.243902 |
The slopes become more negative as the intervals shrink: \(-0.200000,\ -0.227273,\ -0.238095,\ -0.243902\). The values appear to settle near \(-0.25\). To check the calculations exactly, simplify the difference quotient:
For \(h=0.5\), this gives \(-1/[2(2.5)]=-1/5=-0.2\). For \(h=0.2\), it gives \(-1/[2(2.2)]=-1/4.4\approx-0.227273\). For \(h=0.1\), it gives \(-1/4.2\approx-0.238095\), and for \(h=0.05\), it gives \(-1/4.1\approx-0.243902\). These match the table. As another check, multiplying each slope by its interval width recovers the output change; for example, \((-0.238095)(0.1)\approx-0.023810\).
The exact expression also explains why the table points toward \(-0.25\). For positive \(h\), the difference between the slope and \(-1/4\) is
This difference is positive, and it gets smaller for the listed shrinking widths. Thus the slopes are above \(-0.25\) and move closer to it in this example. The table displays the pattern; the algebra checks it for the chosen formula.
How to Build and Read a Table
A reliable table begins with a fixed input \(a\) and a clearly stated function \(f\). Choose several nonzero interval widths \(h\), each smaller than the previous one. For every row, evaluate \(f(a+h)\) and \(f(a)\), subtract in that order, and divide by \(h\). Keeping the subtraction order consistent matters: reversing the output difference without also reversing the input difference changes the sign.
Choose \(a\), and calculate \(f(a)\) once.
Use nonzero values such as \(0.5,\ 0.1,\ 0.01\), and calculate each endpoint \(a+h\).
Find \(f(a+h)-f(a)\), then divide by \(h\).
Look at whether the values move toward an apparent stable number, while noting the displayed precision.
A stable-looking value is a candidate suggested by the entries, not automatically an exact answer. A table might conceal changes beyond its displayed digits, and finitely many rows cannot show what happens at every smaller width. When an exact formula is available, it can help check and explain the numerical evidence. When only measured or rounded data are available, the conclusion should stay appropriately cautious.
Worked Example: A Positive Slope Near \(4\)
Worked Example: A Positive Slope Near \(4\)
Let \(f(x)=x^2\), fix \(a=2\), and use widths \(0.5,\ 0.1,\ 0.01,\ 0.001\). Since \(f(2)=4\), each row compares the value at \(2+h\) with \(4\).
| \(h\) | \(f(2+h)\) | Output change | Secant slope |
|---|---|---|---|
| 0.5 | \(2.5^2=6.25\) | 6.25 − 4 = 2.25 | 2.25 ÷ 0.5 = 4.5 |
| 0.1 | \(2.1^2=4.41\) | 4.41 − 4 = 0.41 | 0.41 ÷ 0.1 = 4.1 |
| 0.01 | \(2.01^2=4.0401\) | 4.0401 − 4 = 0.0401 | 0.0401 ÷ 0.01 = 4.01 |
| 0.001 | \(2.001^2=4.004001\) | 4.004001 − 4 = 0.004001 | 0.004001 ÷ 0.001 = 4.001 |
The entries suggest a stable value near \(4\), with the listed slopes just above it. The exact calculation checks every row:
Substituting the four widths into \(4+h\) gives \(4.5,\ 4.1,\ 4.01,\ 4.001\), exactly the table's slopes. For instance, multiplying \(4.01\) by \(0.01\) gives \(0.0401\), the listed output change. This verifies the quotient's arithmetic and shows why the entries move toward the candidate value in this example.
Worked Example: A Square-Root Table
Worked Example: A Square-Root Table
Now take \(f(x)=\sqrt{x}\), set \(a=4\), and choose \(h=0.5,\ 0.1,\ 0.01\). Here \(f(4)=2\). The output values and slopes below are rounded to six decimal places only after the calculations.
| \(h\) | \(\sqrt{4+h}\) | Output change | Secant slope |
|---|---|---|---|
| 0.5 | 2.121320 | 2.121320 − 2 = 0.121320 | 0.121320 ÷ 0.5 ≈ 0.242641 |
| 0.1 | 2.024846 | 2.024846 − 2 = 0.024846 | 0.024846 ÷ 0.1 ≈ 0.248457 |
| 0.01 | 2.002498 | 2.002498 − 2 = 0.002498 | 0.002498 ÷ 0.01 ≈ 0.249844 |
For example, \(\sqrt{4.5}\approx2.1213203436\), so the unrounded output change is approximately \(0.1213203436\), and dividing by \(0.5\) gives approximately \(0.2426406872\). The analogous unrounded slopes for \(h=0.1\) and \(h=0.01\) are approximately \(0.2484567313\) and \(0.2498439450\). The displayed values are rounded to six decimal places.
There is an algebraic check. Since \(h>0\), the denominator below is nonzero, and multiplying numerator and denominator by \(\sqrt{4+h}+2\) gives
For \(h=0.5\), the last expression is \(1/(\sqrt{4.5}+2)\approx1/4.1213203436\approx0.2426406872\), matching the first slope. For \(h=0.1\), it is \(1/4.0248456731\approx0.2484567313\); for \(h=0.01\), it is \(1/4.0024984395\approx0.2498439450\). Multiplying these unrounded slopes by their widths recovers the corresponding output changes: for example, \(0.2484567313(0.1)\approx0.0248456731\). The table again suggests a stable value, here near \(0.25\).
A Common Sign Mistake
Negative slopes can be misread because “more negative” and “smaller in magnitude” are not the same thing. A tempting but incorrect description of the reciprocal table is that its slopes become less negative as the widths shrink. But its entries go from \(-0.200000\) to \(-0.227273\) to \(-0.238095\). Each new number is numerically smaller, and its absolute value is larger.
A separate sign error can occur inside a row. For \(h=0.2\), the output change is \(f(2.2)-f(2)\approx0.454545-0.5=-0.045455\). Because the input change is \(+0.2\), the quotient is \(-0.045455/0.2\approx-0.227273\). Dividing by \(-0.2\) would incorrectly give a positive slope: the denominator must be the same input change used to define the row.
What the Numerical Evidence Tells You
The secant slopes in these tables are average rates across particular intervals. Their units, when the inputs and outputs represent measured quantities, are output units per input unit. For example, if the input is time in seconds and the output is position in metres, a secant slope is measured in metres per second. The interval widths must be compared on the same input scale, and rounded measurements may limit how much detail a table can reveal.
When entries cluster near a number, it is reasonable to report that the table suggests that number as a candidate stable value near the chosen input. It is not justified to claim that a few rounded entries prove the slopes settle there. The exact algebra in the examples provided a further check because each function had a formula simple enough to simplify. The numerical method is still valuable: it makes the pattern visible, helps catch arithmetic or sign errors, and guides a more exact investigation.
Use a consistent subtraction order, retain enough digits during calculation, and describe the trend accurately. In particular, for negative values, compare both their order on the number line and their distance from the candidate value. Those habits make a shrinking-interval table informative rather than misleading.
Check Your Understanding
Answer the questions using the definitions and examples above.
- For \(f(x)=x^2\), \(a=3\), and \(h=0.1\), write the secant-slope quotient and calculate it.
- In the reciprocal example, why is the slope for \(h=0.2\) negative?
- What does a table of finitely many shrinking intervals suggest, and what does it not prove by itself?
- For \(f(x)=\sqrt{x}\) at \(a=4\), what algebraic expression checks the computed secant slope?
- Explain why the sequence \(-0.20,\ -0.23,\ -0.24\) is becoming more negative, even though the values appear to get closer to \(-0.25\).
