From Change Across an Interval to Change at an Instant
A derivative describes how a function changes at a particular input. A secant slope describes how much the function changes across an interval. Comparing these two quantities leads to a useful question: does the instantaneous rate at some interior point equal the average rate over the whole interval?
For example, if a car’s position changes from 10 meters to 70 meters during a 5-second interval, its average velocity over that interval is 12 meters per second. The car’s velocity may vary during those 5 seconds. It is natural to ask whether, at some moment inside the interval, its instantaneous velocity was exactly 12 meters per second.
The same question can be asked of any differentiable function. The calculation begins with the slope of a secant line and compares it with derivative values at interior points.
The Secant Slope
Let \(f\) be a function defined at two distinct inputs \(a\) and \(b\). The graph contains the points \((a,f(a))\) and \((b,f(b))\). The straight line through these two points is called a secant line to the graph. Its slope is the change in the function values divided by the change in the inputs.
When \(a<b\), we often describe this as the secant slope over \([a,b]\). The interval is closed because both endpoint values are used in the calculation. The formula is the familiar slope formula: vertical change divided by horizontal change.
The secant slope is an average, not necessarily the rate at every point in the interval. If \(f\) measures distance and the input measures time, then this slope has units of distance per time and represents average velocity. More generally, its units are the units of \(f\) divided by the units of its input.
How the Derivative Enters
At an interior point \(c\), where the derivative exists, \(f'(c)\) is the slope of the tangent line to the graph at \((c,f(c))\). It is also the limit of nearby secant slopes as the second input approaches \(c\):
The two slopes answer different questions. The secant slope summarizes the change from one endpoint to another. The derivative describes the local rate at one input. To connect them, set the derivative equal to the secant slope and solve for an interior input \(c\):
This equation is the computational question at the center of the Mean Value Theorem. The theorem will supply conditions under which a suitable interior point is guaranteed to exist. Here, the goal is to understand the two slopes being compared and how to locate such points in particular examples.
Why the Comparison Is Natural
The secant line and the tangent line are both described by slopes. The secant line joins the graph’s endpoint values; the tangent line describes the graph’s direction at a single point. If their slopes are equal, the tangent line is parallel to the secant line.
This parallelism can also be expressed without relying on a sketch. The line through \((a,f(a))\) and \((b,f(b))\) has slope \(\frac{f(b)-f(a)}{b-a}\). A tangent at \(c\) has slope \(f'(c)\). Equal slopes mean the two lines have the same direction, even though they generally pass through different points.
The comparison is meaningful when the function is defined at the endpoints, so the secant slope can be calculated, and when it has derivatives at the candidate interior points. A theorem giving a general existence guarantee requires precise hypotheses; checking those hypotheses is a separate task. The examples below use functions that are differentiable throughout the intervals in question.
Worked Examples
Worked Example: A Quadratic on an Interval
Let \(f(x)=x^2\) on \([1,3]\). First calculate the endpoint values:
The change in output is \(9-1=8\), and the change in input is \(3-1=2\). Therefore the secant slope is
Differentiate \(f(x)=x^2\): \(f'(x)=2x\). Set the derivative equal to the secant slope:
Since \(1<2<3\), this is an interior point. Check the calculation by substituting it into the derivative: \(f'(2)=2(2)=4\), which matches the secant slope. Geometrically, the tangent at \(x=2\) is parallel to the secant joining the graph’s endpoints.
Worked Example: More Than One Matching Point
Let \(f(x)=x^3\) on \([-1,2]\). The endpoint values are
Thus the secant slope is
The derivative is \(f'(x)=3x^2\). We solve \(f'(c)=3\):
The required input must be interior to \((-1,2)\). The value \(-1\) is an endpoint, so it is not an interior solution; \(1\) is interior. Checking gives \(f'(1)=3(1)^2=3\), equal to the secant slope. This example illustrates why the interior condition matters and why there can be more than one algebraic solution before that condition is applied.
Worked Example: A Reciprocal Function
Let \(f(x)=\frac{1}{x}\) on \([1,2]\). The endpoint values are \(f(1)=1\) and \(f(2)=\frac12\). The secant slope is
Differentiate \(f(x)=x^{-1}\): \(f'(x)=-x^{-2}=-\frac{1}{x^2}\). Solve for \(c\):
The solutions to this equation are \(c=\sqrt{2}\) and \(c=-\sqrt{2}\). Only \(\sqrt{2}\) lies in \((1,2)\), since \(1<\sqrt{2}<2\). Check by substitution:
This equals the secant slope, as required.
A Short Derivation: Average Rate and Secant Slope
The secant-slope formula follows directly from the definition of average rate. Suppose the input changes from \(a\) to \(b\). The total change in the output is \(f(b)-f(a)\), while the total change in the input is \(b-a\). Average rate is defined as total output change divided by total input change. Therefore,
This is also precisely the slope formula for the line joining the points \((a,f(a))\) and \((b,f(b))\). Thus the algebraic average rate and the geometric secant slope are two descriptions of the same quantity.
The derivative is built from the same slope formula, but with one endpoint fixed and the other moving closer to it. At \(c\), the nearby secant slope is \(\frac{f(c+h)-f(c)}{h}\); taking its limit as \(h\) approaches zero gives \(f'(c)\). The comparison between the average rate and the derivative is therefore a comparison between a finite-interval slope and a limiting local slope.
A Common Mistake: Assuming the Matching Point Is the Midpoint
A tempting shortcut is to assume the derivative at the interval’s midpoint must equal the secant slope. That is not generally true. For \(f(x)=x^3\) on \([0,2]\), the midpoint is \(1\). The endpoint values are \(f(0)=0\) and \(f(2)=8\), so the secant slope is
But \(f'(x)=3x^2\), and the derivative at the midpoint is \(f'(1)=3\), not \(4\). The midpoint does not solve the equation in this example. Solving it correctly gives
Only \(c=\frac{2}{\sqrt{3}}\) lies in \((0,2)\). Substitution verifies \(f'(c)=3\left(\frac{2}{\sqrt{3}}\right)^2=3\left(\frac43\right)=4\). The correct method is to calculate the secant slope, solve the derivative equation, and then check which solutions lie strictly inside the interval.
What This Comparison Tells Us
The secant slope compresses all the change across an interval into one number. The derivative gives a rate at a particular input. Finding an interior point where they agree identifies an instant when the local rate matches the average rate across the full interval.
In applications, this can turn a statement about an overall change into a statement about a rate at some moment. In graphing and analysis, it connects the direction of a chord to the direction of a tangent. The Mean Value Theorem formalizes the general existence question, under its stated conditions; the examples here show what the equation means and how its solutions are found.
Check Your Understanding
For each question, focus on the distinction between the interval-wide secant slope and a derivative at one input.
- What information does the secant slope \(\frac{f(b)-f(a)}{b-a}\) summarize?
- For \(f(x)=x^2\) on \([0,4]\), calculate the secant slope and find an interior \(c\) satisfying \(f'(c)\) equal to that slope.
- Why does a solution of \(f'(c)=\frac{f(b)-f(a)}{b-a}\) need to be checked against \(a<c<b\)?
- Can there be more than one interior point where the derivative equals the secant slope? Explain using an example from the tutorial.
- What is the geometric relationship between the secant line and a tangent line when their slopes are equal?
