From a Secant Slope to a Derivative
A secant line through two points on a function’s graph records the function’s average rate of change between those points. The Mean Value Theorem says that, under suitable conditions, the graph has an interior point where its tangent slope equals that secant slope.
Rolle’s Theorem is a special case in which a function has equal values at the two endpoints and therefore has a horizontal tangent somewhere between them. The key to deriving the Mean Value Theorem is to subtract the secant line from the original function. The difference has equal endpoint values, so Rolle’s Theorem applies.
The Mean Value Theorem
Let \(f\) be a function defined on the closed interval \([a,b]\), where \(a<b\). Its secant slope between the endpoints is
This quotient is the average rate of change of \(f\) over \([a,b]\). The Mean Value Theorem guarantees that this average rate is attained as an instantaneous rate at some interior point, provided \(f\) is continuous on the closed interval and differentiable on the open interval.
The conclusion concerns a point strictly inside the interval. The theorem does not say that every interior point has this derivative, nor does it require the derivative to equal the average rate at an endpoint.
Constructing the Auxiliary Function
The straight line through the points \((a,f(a))\) and \((b,f(b))\) has slope \(\frac{f(b)-f(a)}{b-a}\). Writing it in point-slope form gives the secant line
Now define the difference between the function and this line:
At both endpoints, the function and its secant line have the same values. Therefore \(g(a)=g(b)=0\). Also, subtracting a line preserves continuity and differentiability wherever \(f\) has those properties. Thus \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\).
Rolle’s Theorem applies to \(g\): there is some \(c\in(a,b)\) such that \(g'(c)=0\). Differentiating the definition of \(g\) then gives the Mean Value Theorem’s conclusion.
Derivation from Rolle’s Theorem
Here is the complete argument. The only theorem needed for the key existence step is Rolle’s Theorem: if a function is continuous on \([a,b]\), differentiable on \((a,b)\), and has equal endpoint values, then its derivative is zero at some point of \((a,b)\).
Set \(m=\frac{f(b)-f(a)}{b-a}\), and define \(L(x)=f(a)+m(x-a)\). This line passes through \((a,f(a))\) and \((b,f(b))\).
Define \(g(x)=f(x)-L(x)\). Since \(L(a)=f(a)\), \(g(a)=0\). Since \(L(b)=f(a)+m(b-a)=f(b)\), \(g(b)=0\).
The assumptions on \(f\) imply that \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Because \(g(a)=g(b)\), Rolle’s Theorem gives a \(c\in(a,b)\) with \(g'(c)=0\).
Since \(L'(x)=m\), differentiating \(g=f-L\) yields \(g'(x)=f'(x)-m\). Hence \(0=g'(c)=f'(c)-m\), so \(f'(c)=\frac{f(b)-f(a)}{b-a}\), as required.
Every condition has a role in this argument. Continuity on the closed interval and differentiability on the open interval are needed to use Rolle’s Theorem on \(g\). The condition \(a<b\) ensures that the secant slope is defined and that there is an open interval in which the point \(c\) can lie.
Why the Secant Line Is the Natural Choice
The secant line makes the auxiliary function vanish at both endpoints, which makes the application of Rolle’s Theorem especially transparent. Its slope is also exactly the target value in the Mean Value Theorem. Indeed, differentiating \(g=f-L\) subtracts that slope from \(f'\), so a point where \(g'=0\) is precisely a point where \(f'\) equals the secant slope.
The intercept matters for making the endpoint values zero, but it is important to be precise about what the proof requires. Any line with the same slope as the secant line produces an auxiliary function with equal endpoint values; the two values need not be zero. Rolle’s Theorem requires equal endpoint values, not zero endpoint values. The secant line is chosen because it gives the convenient values \(g(a)=g(b)=0\).
Here \(m=\frac{f(b)-f(a)}{b-a}\), while \(k\) can be any constant. Thus \(h(a)=h(b)\) for every such parallel line. Taking \(k=f(a)-ma\) makes \(P\) the secant line and gives \(h(a)=h(b)=0\).
Worked Examples
Worked Example: Deriving the Mean Value Point for a Quadratic
Let \(f(x)=x^2\) on \([1,3]\). Find the secant line, construct the auxiliary function, and verify the point supplied by the derivation.
First check the conditions: a polynomial is continuous and differentiable for every real \(x\), so \(f\) is continuous on \([1,3]\) and differentiable on \((1,3)\).
The endpoint values are \(f(1)=1\) and \(f(3)=9\). The secant slope is
The secant line through \((1,1)\) with slope \(4\) is \(L(x)=1+4(x-1)=4x-3\). Therefore
Check the endpoint values: \(g(1)=1-4+3=0\), and \(g(3)=9-12+3=0\). Differentiating gives \(g'(x)=2x-4\). Setting \(g'(c)=0\) gives \(2c-4=0\), so \(c=2\), which lies in \((1,3)\).
Finally, check the Mean Value Theorem equation directly: \(f'(x)=2x\), so \(f'(2)=4\), equal to the secant slope \(4\). The constructed function also checks: \(g'(2)=2(2)-4=0\).
Worked Example: A Cubic with Equal Endpoint Values After Subtraction
Let \(f(x)=x^3\) on \([-1,2]\). Use the secant-line construction to find an interior point where the derivative equals the average rate of change.
Because \(f\) is a polynomial, it is continuous on \([-1,2]\) and differentiable on \((-1,2)\). Its endpoint values are \(f(-1)=-1\) and \(f(2)=8\). Thus the secant slope is
The line through \((-1,-1)\) with slope \(3\) is \(L(x)=-1+3(x+1)=3x+2\). Subtracting it gives
At the endpoints, \(g(-1)=-1+3-2=0\) and \(g(2)=8-6-2=0\). Differentiate: \(g'(x)=3x^2-3\). Solving \(g'(c)=0\) gives \(3c^2-3=0\), so \(c^2=1\), and \(c=-1\) or \(c=1\). Only \(c=1\) is in the open interval \((-1,2)\).
Check the answer using \(f'(x)=3x^2\): \(f'(1)=3\), which equals the secant slope. Also \(g'(1)=3-3=0\), confirming the Rolle point in the auxiliary function.
Worked Example: Applying the Construction to a Radical Function
Let \(f(x)=\sqrt{x}\) on \([0,4]\). Find the point given by the Mean Value Theorem and check it.
The square-root function is continuous on \([0,4]\). For every \(x\in(0,4)\), it is differentiable, with \(f'(x)=\frac{1}{2\sqrt{x}}\). Therefore the theorem’s hypotheses hold.
The endpoint values are \(f(0)=0\) and \(f(4)=2\), so the secant slope is
The secant line through \((0,0)\) with slope \(\frac12\) is \(L(x)=\frac{x}{2}\). The difference function is \(g(x)=\sqrt{x}-\frac{x}{2}\). Its endpoints are \(g(0)=0-\frac{0}{2}=0\) and \(g(4)=2-\frac{4}{2}=0\).
For \(x\in(0,4)\), differentiate to obtain
Set this equal to zero: \(\frac{1}{2\sqrt{c}}-\frac12=0\), so \(\frac{1}{2\sqrt{c}}=\frac12\), then \(\sqrt{c}=1\), and \(c=1\). This value lies in \((0,4)\). Checking directly, \(f'(1)=\frac{1}{2\sqrt{1}}=\frac12\), exactly the secant slope; also \(g'(1)=\frac12-\frac12=0\).
Common Pitfall: Confusing Equal Values with Zero Values
Consider \(f(x)=x^2\) on \([1,3]\), whose secant slope is \(4\). The secant line is \(4x-3\), so subtracting it gives endpoint values of zero. But suppose instead we subtract the parallel line \(P(x)=4x+5\). Then
Here \(h(1)=1-4-5=-8\), while \(h(3)=9-12-5=-8\). A mistaken conclusion would be: “The endpoint values are not zero, so Rolle’s Theorem cannot be used.” That is incorrect. Rolle’s Theorem requires \(h(1)=h(3)\), and both values are \(-8\). The function \(h\) is continuous on \([1,3]\) and differentiable on \((1,3)\), so Rolle’s Theorem applies. Indeed, \(h'(x)=2x-4\), and \(h'(2)=0\). Since \(h'=f'-4\), this gives \(f'(2)=4\), the secant slope.
Why This Derivation Matters
The proof is more than an algebraic trick. It shows how an existence statement about a horizontal tangent can establish an existence statement about a tangent with any specified slope. Subtracting the secant line removes the slope being targeted: a tangent to \(f\) parallel to the secant line becomes a horizontal tangent to \(g=f-L\).
The construction also clarifies the logic of the theorem. The secant line sets the target slope, matching endpoint values let Rolle’s Theorem apply, and differentiating the difference converts the zero derivative into the required equality for \(f'\). Each part depends on the stated continuity and differentiability hypotheses; without them, this derivation cannot invoke Rolle’s Theorem as written.
Check Your Understanding
For each question, focus on how the auxiliary function connects the secant slope to Rolle’s Theorem.
- Write the secant line through \((a,f(a))\) and \((b,f(b))\) using point-slope form.
- For \(g(x)=f(x)-L(x)\), explain why \(g(a)=g(b)=0\) when \(L\) is the secant line.
- If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), which hypotheses allow Rolle’s Theorem to be applied to \(g\)?
- Why does \(g'(c)=0\) imply \(f'(c)=\frac{f(b)-f(a)}{b-a}\)?
- If a line has the secant slope but a different intercept, must the corresponding difference function have zero endpoint values? What endpoint property does it still have?
