Why Combine the Two Theorems?
On a closed interval, continuity guarantees that a function attains an absolute maximum and an absolute minimum. If the function is also differentiable in the interval’s interior, the Mean Value Theorem can connect the values at those extremum points to a derivative at an interior point.
The order matters: first identify points in the specified interval where the extreme values occur; then apply the Mean Value Theorem between those points. The resulting derivative conclusion depends on the actual locations of the extrema, not automatically on the full interval’s endpoints.
Precise Statements
In a combined problem, the Extreme Value Theorem (EVT) supplies points where the largest and smallest values are attained. The Mean Value Theorem (MVT) can then be applied between those points, provided the function is continuous on the interval between them and differentiable in its interior.
A Combined Consequence
Suppose \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Write \(M\) for its absolute maximum value and \(m\) for its absolute minimum value. If \(f\) is not constant, then \(M>m\), so a point attaining the maximum and a point attaining the minimum must be distinct. Choose such points and name their coordinates \(u\) and \(v\), ordered so that \(u<v\). Applying the MVT on \([u,v]\) gives an interior point where the derivative equals the secant slope between the extremum values at those coordinates.
The derivative’s sign depends on which extremum occurs first along the interval: the slope is positive if the function rises from the minimum to the maximum, and negative if it falls from the maximum to the minimum. The inequality follows because \(0<v-u\leq b-a\), so dividing the positive range \(M-m\) by \(v-u\) gives at least as large a result as dividing it by \(b-a\).
Why the Combined Conclusion Holds
Here is the full argument. By the EVT, there are points \(x_{\max},x_{\min}\in[a,b]\) where \(f\) attains \(M\) and \(m\). Since \(f\) is nonconstant, \(M>m\), and the two points cannot be the same. Order their coordinates as \(u<v\). The function is continuous on \([u,v]\), because it is continuous on \([a,b]\), and differentiable on \((u,v)\), because \((u,v)\subset(a,b)\). Thus the MVT applies on \([u,v]\).
The two endpoint values on this smaller interval are \(M\) and \(m\), in one order or the other. Consequently, the absolute value of their secant slope is
By the MVT, there is a \(c\in(u,v)\) for which \(f'(c)\) equals that signed secant slope. Taking absolute values yields the stated derivative magnitude. Finally, \(v-u\leq b-a\), so
This combines the roles of the theorems without confusing them: the EVT guarantees that the extreme values occur, and the MVT guarantees a matching derivative at an interior point between the points where those values occur.
Worked Applications
Worked Example: Find the Extrema, Then Apply the MVT
Let \(f(x)=x^2-2x\) on \([-1,3]\). Find its absolute extrema, then use the MVT between a point attaining the maximum and a point attaining the minimum.
Because \(f\) is a polynomial, it is continuous on \([-1,3]\). The EVT guarantees that it attains both absolute extrema on this interval. To identify them, check the endpoints and the interior point where \(f'(x)=0\):
The candidate values are
Thus the absolute maximum is \(3\), attained at both \(-1\) and \(3\), and the absolute minimum is \(-1\), attained at \(1\). Choose the maximum point \(-1\) and the minimum point \(1\). The function is continuous on \([-1,1]\) and differentiable on \((-1,1)\), so the MVT gives a \(c\in(-1,1)\) such that
Solve using the derivative:
Check: \(0\in(-1,1)\), and \(f'(0)=2(0)-2=-2\), which matches the secant slope. The range is \(3-(-1)=4\), while the distance between the chosen extremum points is \(2\), so the derivative magnitude is \(4/2=2\).
Worked Example: An Endpoint Minimum and an Interior Maximum
Let \(g(x)=x(2-x)\) on \([0,2]\). Determine its absolute extrema and find a Mean Value point between an endpoint minimum and the maximum.
The function is a polynomial, so it is continuous on the closed interval and the EVT guarantees attained extrema. Expand and differentiate:
The interior critical point satisfies \(2-2x=0\), so \(x=1\). The endpoint and critical-point values are
Therefore, the absolute maximum is \(1\) at \(x=1\), and the absolute minimum is \(0\) at both endpoints. Choose the minimum point \(0\) and maximum point \(1\). The MVT applies on \([0,1]\), so some \(c\in(0,1)\) satisfies
Now solve:
Check: \(\frac12\in(0,1)\), and \(g'(\frac12)=2-2(\frac12)=1\), the required slope. This example also shows that an extremum point may be an endpoint; the combined argument uses the interval between the selected extremum points.
Worked Example: Multiple Extrema and More Than One MVT Point
Let \(h(x)=x^3-3x\) on \([-2,2]\). Find its absolute extrema, then determine all MVT points between one maximum point and one minimum point.
The polynomial is continuous on \([-2,2]\), so the EVT guarantees that it attains a maximum and minimum. Its derivative is
The interior critical points are \(x=-1\) and \(x=1\). Evaluate at the endpoints and critical points:
The absolute maximum is \(2\), attained at \(-1\) and \(2\); the absolute minimum is \(-2\), attained at \(-2\) and \(1\). Choose the maximum point \(-1\) and minimum point \(1\). The MVT on \([-1,1]\) guarantees a \(c\in(-1,1)\) with
Solve the derivative equation:
Both values lie in \((-1,1)\). For either value, \(h'(c)=3(\frac13)-3=1-3=-2\), so both satisfy the MVT equation. The theorem guarantees at least one such point; it does not say that the point must be unique.
Common Pitfalls
An extremum on a specified interval must be attained at a point in that interval. The function may have a larger value elsewhere in its domain, but that value is irrelevant to the maximum on the interval.
For example, consider \(f(x)=x\) on \([0,1]\). The following is an incorrect argument: “For every \(x\in[0,1]\), \(f(x)\leq f(2)=2\), so the maximum is \(2\).” The inequality is true, but \(2\notin[0,1]\); \(f(2)\) is not attained on the interval. The actual maximum is \(f(1)=1\), and the minimum is \(f(0)=0\).
A second error is applying the MVT across the full interval when the chosen maximum and minimum occur at interior points. The theorem’s secant slope must use the two points that bound the interval on which it is applied. In the first example, the selected points are \(-1\) and \(1\), so the relevant slope is \(\frac{f(1)-f(-1)}{1-(-1)}=-2\), not a slope computed from some other pair of points.
Finally, the EVT and MVT have different hypotheses. Continuity on a closed interval gives attained extrema but does not, by itself, allow the MVT. To use the MVT between the extremum points, the function must also be differentiable in the open interval between them. If that condition fails, the combined derivative conclusion has not been established.
Check Your Understanding
For each question, identify which theorem supplies the needed conclusion and check the relevant interval.
- What conditions ensure that a function attains both an absolute maximum and an absolute minimum on \([a,b]\)?
- If a nonconstant function has absolute maximum \(M\) and minimum \(m\) on \([a,b]\), what derivative-magnitude bound follows from combining the EVT and MVT?
- For \(f(x)=x^2\) on \([-1,2]\), find the absolute extrema and then find an MVT point between a minimum point and a maximum point.
- Why is \(x_{\max}\in[a,b]\) part of the definition of an absolute maximum on \([a,b]\)?
- Does the EVT alone guarantee a point where \(f'(c)\) equals a secant slope? State the additional condition needed.
