From a Secant Slope to a Tangent Slope
A secant line records the average change of a function between two points. The Mean Value Theorem says that, under suitable conditions, this average rate of change is attained as an instantaneous rate of change somewhere between the endpoints. The geometric meaning is that at some interior point, the tangent line is parallel to the secant line.
Rolle’s Theorem handles a special case: when the two endpoint values of a function are equal, its derivative is zero at some interior point. To derive the Mean Value Theorem, the central task is to transform the original function into one whose endpoint values are equal, without losing the relationship to its derivative.
The Statement and the Construction
The quotient on the right is the slope of the secant line through \((a,f(a))\) and \((b,f(b))\). Denote this slope by \(m\). The equation of the line through those two points is \(L(x)=f(a)+m(x-a)\). Subtracting \(L\) from \(f\) gives an auxiliary function whose values at \(a\) and \(b\) agree.
The construction is chosen for two linked reasons. First, the line passes through both endpoints of the graph, so subtracting it makes the auxiliary function zero at both endpoints. Second, the line has constant derivative \(m\), so the derivative of the auxiliary function is \(f'(x)-m\). Rolle’s Theorem can therefore force \(f'(x)-m\) to equal zero at an interior point.
Derivation from Rolle’s Theorem
We use Rolle’s Theorem in its standard form: if a function is continuous on \([a,b]\), differentiable on \((a,b)\), and has equal values at \(a\) and \(b\), then its derivative is zero at some \(c\in(a,b)\). The proof below constructs a function satisfying exactly these requirements.
Since \(a<b\), the denominator \(b-a\) is positive, so the number \(m=\frac{f(b)-f(a)}{b-a}\) is defined.
Define \(g(x)=f(x)-[f(a)+m(x-a)]\) for \(x\in[a,b]\). The function \(g\) is continuous on \([a,b]\) and differentiable on \((a,b)\), because \(f\) has those properties and the subtracted function is linear.
At \(a\), \(g(a)=f(a)-f(a)=0\). At \(b\), \(g(b)=f(b)-f(a)-m(b-a)=0\), using \(m(b-a)=f(b)-f(a)\). Thus \(g(a)=g(b)\).
There is a \(c\in(a,b)\) such that \(g'(c)=0\).
Since \(g'(x)=f'(x)-m\), the equality \(g'(c)=0\) gives \(f'(c)=m=\frac{f(b)-f(a)}{b-a}\), as required.
This proves the Mean Value Theorem. The argument does not assume that \(f\) is linear, nor that the point \(c\) is unique. It establishes the existence of at least one interior point with the specified derivative.
Worked Examples of the Auxiliary Function
Worked Example: A Quadratic on an Interval
Let \(f(x)=x^2\) on \([1,3]\). The endpoint values are \(f(1)=1\) and \(f(3)=9\), so the secant slope is \(\frac{9-1}{3-1}=\frac{8}{2}=4\). The secant line through \((1,1)\) is \(L(x)=1+4(x-1)=4x-3\).
Subtract the line: \(g(x)=x^2-(4x-3)=x^2-4x+3=(x-1)(x-3)\). Thus \(g(1)=0\) and \(g(3)=0\). Since \(g\) is a polynomial, it is continuous on \([1,3]\) and differentiable on \((1,3)\). Rolle’s Theorem gives an interior \(c\) with \(g'(c)=0\).
Differentiate: \(g'(x)=2x-4\). Solving \(2c-4=0\) gives \(c=2\), which lies in \((1,3)\). Since \(g'(x)=f'(x)-4\), this gives \(f'(2)=4\). Directly, \(f'(x)=2x\), so \(f'(2)=2(2)=4\), matching the secant slope. The construction and conclusion agree.
Worked Example: A Linear Function
Let \(f(x)=3x+2\) on \([0,2]\). Its endpoint values are \(f(0)=2\) and \(f(2)=8\). The secant slope is \(\frac{8-2}{2-0}=\frac{6}{2}=3\), and the secant line is \(L(x)=2+3(x-0)=3x+2\).
The auxiliary function is \(g(x)=f(x)-L(x)=(3x+2)-(3x+2)=0\) for every \(x\in[0,2]\). Therefore \(g(0)=g(2)=0\), and \(g'(x)=0\) throughout \((0,2)\). Every \(c\in(0,2)\) satisfies the conclusion.
To check the Mean Value Theorem directly, \(f'(x)=3\) for every \(x\), while the secant slope is also \(3\). Hence \(f'(c)=\frac{f(2)-f(0)}{2-0}=3\) for any interior \(c\). This example shows that the auxiliary function need not have just one point where its derivative is zero.
Worked Example: A Cubic with Two Endpoint Zeros
Let \(f(x)=x^3\) on \([-1,2]\). We have \(f(-1)=-1\) and \(f(2)=8\). The secant slope is \(\frac{8-(-1)}{2-(-1)}=\frac{9}{3}=3\). The secant line through \((-1,-1)\) is \(L(x)=-1+3(x+1)=3x+2\).
Subtracting gives \(g(x)=x^3-(3x+2)=x^3-3x-2\). Factoring, \(g(x)=(x-2)(x+1)^2\). Consequently \(g(-1)=0\) and \(g(2)=0\). This polynomial is continuous on the closed interval and differentiable on its interior, so Rolle’s Theorem applies.
Differentiate \(g\): \(g'(x)=3x^2-3\). At \(c=1\), which lies in \((-1,2)\), \(g'(1)=3(1)^2-3=0\). Since \(f'(x)=3x^2\), we obtain \(f'(1)=3\), equal to the secant slope \(3\). The derivative relation also checks directly: \(g'(1)=f'(1)-3=3-3=0\).
Why the Line Must Be the Secant Line
Subtracting any line changes a function’s derivative by a constant, but the chosen line must also make the endpoint values of the difference equal. The secant line is precisely the line that meets the graph at both endpoints. Its slope is fixed by the endpoint data, and that slope is the quantity the Mean Value Theorem asserts will occur as a derivative.
For the auxiliary function \(g=f-L\), the two parts of the proof therefore fit together: the endpoint geometry gives \(g(a)=g(b)\), while the derivative calculation gives \(g'=f'-m\). Rolle’s Theorem supplies a point where the latter expression is zero. If the subtracted line had a different slope, a zero of \(g'\) would instead show that \(f'\) equals that different slope, which would not establish the theorem’s required conclusion.
Common Pitfall: Subtracting Only a Constant
A tempting but usually incorrect construction is \(g(x)=f(x)-f(a)\). It makes \(g(a)=0\), but does not generally make \(g(b)=0\). Equal endpoint values are essential before Rolle’s Theorem can be applied.
Worked Example: Correcting an Incomplete Construction
Take \(f(x)=x^2\) on \([1,3]\). Subtracting only \(f(1)=1\) gives \(q(x)=x^2-1\). Its endpoint values are \(q(1)=1-1=0\) and \(q(3)=9-1=8\). Since \(q(1)\ne q(3)\), Rolle’s Theorem does not apply to \(q\).
The correction is to subtract the whole secant line. The secant slope is \((9-1)/(3-1)=4\), so \(L(x)=1+4(x-1)=4x-3\). Now \(g(x)=x^2-(4x-3)=x^2-4x+3\), with \(g(1)=1-4+3=0\) and \(g(3)=9-12+3=0\). Thus the corrected auxiliary function has equal endpoint values, and its derivative \(g'(x)=2x-4\) gives the required relation \(g'(c)=f'(c)-4=0\).
Another error is to omit the derivative of the line. Because \(L'(x)=m\), differentiating \(f(x)-L(x)\) gives \(f'(x)-m\), not just \(f'(x)\). The subtraction is what converts the conclusion \(g'(c)=0\) into the specific equality \(f'(c)=m\).
What the Proof Establishes
The proof is a general mechanism: start with a function whose endpoint changes need not be zero, remove the linear change represented by its secant line, and apply Rolle’s Theorem to what remains. The continuity and differentiability assumptions pass from \(f\) to the auxiliary function because a linear function is continuous and differentiable everywhere. The endpoint equality follows by direct substitution, and the derivative conclusion follows by differentiation.
Each condition has a role in the argument. The inequality \(a<b\) ensures that the secant slope has a nonzero denominator. Continuity on the closed interval and differentiability on the open interval are the hypotheses needed to use Rolle’s Theorem on \(g\). The proof does not claim a value of \(c\) at either endpoint; Rolle’s Theorem locates it in \((a,b)\).
Check Your Understanding
Answer each question using the construction and proof above.
- Why is the slope \(m=\frac{f(b)-f(a)}{b-a}\) defined when \(a<b\)?
- Write the equation of the secant line through \((a,f(a))\) and \((b,f(b))\).
- For \(g(x)=f(x)-[f(a)+m(x-a)]\), verify separately that \(g(a)=0\) and \(g(b)=0\).
- Why does \(g\) satisfy the continuity and differentiability requirements for Rolle’s Theorem?
- If Rolle’s Theorem gives \(g'(c)=0\), show how the derivative of the subtracted line yields the Mean Value Theorem’s conclusion.
