From a Secant Slope to a Point
A polynomial’s values at the endpoints of an interval determine the slope of the line joining those endpoint values. The Mean Value Theorem guarantees that somewhere between the endpoints, the polynomial has an instantaneous rate of change equal to that secant slope. For a polynomial, finding such a point becomes a concrete algebra problem: differentiate, set the derivative equal to the secant slope, and solve.
The calculation has two parts that should be kept distinct. First, the theorem guarantees at least one point in the open interval. Then the derivative equation identifies the possible points. Some algebraic solutions may fall outside the interval, and endpoint solutions do not count. We will therefore solve the equation and check the interval condition before reporting the answer.
The Guarantee and the Equation to Solve
For a polynomial \(p\), compute the endpoint values \(p(a)\) and \(p(b)\), then form the secant slope. The points sought are precisely the solutions \(c\) in the open interval \((a,b)\) of
The equation may have algebraic solutions outside \((a,b)\), which the theorem does not identify as Mean Value Theorem points for this interval. It may also have a solution at an endpoint, which must be excluded. The theorem guarantees at least one solution strictly between \(a\) and \(b\), but the equation is what we use to find all the points for the particular polynomial and interval.
Why This Equation Finds the Guaranteed Point
The equation is not an arbitrary recipe: it states that the tangent slope at \(c\) equals the secant slope across the whole interval. One way to see why the equation is appropriate is to subtract the secant line from the polynomial. Let
The line \(p(a)+m(x-a)\) has value \(p(a)\) at \(a\) and value \(p(b)\) at \(b\). Thus \(q(a)=0\) and \(q(b)=0\). Since \(q\) is a polynomial, Rolle’s Theorem applies: for some \(c\in(a,b)\), \(q'(c)=0\). Differentiating the definition of \(q\) gives \(q'(x)=p'(x)-m\). Therefore \(q'(c)=0\) means \(p'(c)=m\), exactly the equation above.
This reasoning explains the guarantee and the slope equation. To calculate the point or points, we now need only evaluate the polynomial at the endpoints, differentiate, solve the resulting equation, and retain the solutions in the interval.
A Reliable Calculation Strategy
Record the endpoints \(a<b\). The answer must lie in the open interval \((a,b)\).
Evaluate \(p(a)\) and \(p(b)\), then calculate \(\bigl(p(b)-p(a)\bigr)/(b-a)\).
Find \(p'(x)\), set it equal to the secant slope, and solve the equation exactly where possible.
Keep only solutions strictly between \(a\) and \(b\), then substitute each retained value into \(p'(x)\) to check the slope.
Showing the endpoint values and the interval check is part of a complete answer. A solution to the derivative equation alone is only a candidate until its location and slope have been confirmed.
Worked Examples
Worked Example: A Quadratic on a Closed Interval
Find all Mean Value Theorem points for \(p(x)=x^2+x\) on \([0,2]\).
Step 1: Check the guarantee. A polynomial is continuous and differentiable everywhere, so the Mean Value Theorem applies on \([0,2]\).
Step 2: Compute the secant slope. The endpoint values are
Since \(b-a=2-0=2\), the secant slope is
Step 3: Solve the derivative equation. Differentiate and set the derivative equal to \(3\):
Step 4: Check the interval and slope. Since \(0<1<2\), the candidate is in the required open interval. Substitution gives \(p'(1)=2(1)+1=3\), equal to the secant slope. Therefore the Mean Value Theorem point is \(c=1\).
Worked Example: A Cubic with Two Points in the Interval
Find all Mean Value Theorem points for \(p(x)=x^3-3x\) on \([-2,2]\).
Step 1: Check the guarantee. The function is a polynomial, so it is continuous on \([-2,2]\) and differentiable on \((-2,2)\).
Step 2: Compute the secant slope.
Thus the secant slope is
Step 3: Solve the derivative equation.
Taking both square roots gives \(x=\frac{2}{\sqrt{3}}\) or \(x=-\frac{2}{\sqrt{3}}\). Both are in \((-2,2)\), since \(\frac{2}{\sqrt{3}}<2\); the same strict bound holds for its negative.
Step 4: Verify both slopes. For either candidate, \(x^2=\frac43\), so
This equals the secant slope. The complete set of points is \(c=-\frac{2}{\sqrt{3}}\) and \(c=\frac{2}{\sqrt{3}}\).
Worked Example: Excluding an Endpoint Solution
Find all Mean Value Theorem points for \(p(x)=x^3-3x^2+x\) on \([0,3]\).
Step 1: Check the guarantee. The polynomial meets the continuity and differentiability conditions on the stated interval.
Step 2: Compute the secant slope.
Therefore, the secant slope is
Step 3: Solve the derivative equation.
Subtract \(1\) from each side and factor:
So the algebraic solutions are \(x=0\) and \(x=2\). The endpoint \(0\) is not in the open interval \((0,3)\), while \(2\) is. Hence only \(c=2\) qualifies. Check it directly:
The derivative at \(2\) equals the secant slope, and \(0<2<3\). Thus the Mean Value Theorem point is \(c=2\); the other equation solution is excluded because it is an endpoint.
Common Pitfall: Confusing Candidates with Answers
A frequent error is to report every solution of the derivative equation without checking whether it lies in the open interval. In the third example, the equation produced \(x=0\) and \(x=2\). Reporting both as Mean Value Theorem points would be incorrect: the theorem requires \(c\in(0,3)\), and \(0\notin(0,3)\).
Another error is to use the wrong secant slope. The numerator must be the change in function value, \(p(b)-p(a)\), and the denominator must be the change in input, \(b-a\). For example, on \([0,2]\) for \(p(x)=x^2+x\), using \(p(2)-p(0)=6\) as the slope would lead to \(2x+1=6\). But the actual secant slope is \(6/(2-0)=3\), so the correct equation is \(2x+1=3\). Always show the denominator before solving.
What the Calculation Establishes
For a polynomial, the procedure turns a theorem’s existence guarantee into an explicit search. The endpoint values set the target slope; the derivative equation finds where the polynomial’s instantaneous slope reaches that target. Checking the open interval makes sure the answer satisfies the theorem’s conclusion for the stated interval.
The final verification is especially useful when solving equations creates several candidates or when an endpoint appears as a root. For each reported value \(c\), confirm both conditions:
- \(a<c<b\), so the point is strictly inside the interval;
- \(p'(c)=\bigl(p(b)-p(a)\bigr)/(b-a)\), so its derivative equals the secant slope.
These checks distinguish a valid Mean Value Theorem point from a merely plausible algebraic result.
Check Your Understanding
For each problem, compute the secant slope, solve the derivative equation, and retain only points in the open interval.
- Find all Mean Value Theorem points for \(p(x)=x^2\) on \([1,3]\).
- For \(p(x)=x^3\) on \([-1,1]\), what secant slope must \(p'(c)\) equal?
- For \(p(x)=x^3-3x^2\) on \([0,3]\), solve the derivative equation and identify which solutions lie in \((0,3)\).
- Why is a solution at \(x=a\) or \(x=b\) not a Mean Value Theorem point for \([a,b]\)?
- What two checks should be made for every reported candidate \(c\)?
