Rolle and the Mean Value Theorem

We can combine our definition of derivative with the Intermediate Value Theorem to give a useful result which is in fact the basis of most elementary applications of the differential calculus. Like the results on continuous functions, it is a global result, and so needs continuity and differentiability on a whole interval.

Theorem 5.15 (Rolle's Theorem)   Let f be continuous on [a, b], and differentiable on (a, b), and suppose that f (a) = f (b). Then there is some c with a < c < b such that f'(c) = 0.

Note:The theorem guarantees that the point c exists somewhere. It gives no indication of how to find c. Here is the diagram to make the point geometrically:

Figure 5.1: If f crosses the axis twice, somewhere between the two crossings, the function is flat. The accurate statement of this ``obvious'' observation is Rolle's Theorem.

Proof. Since f is continuous on the compact interval [a, b], it has both a global maximum and a global minimum. Assume first that the global maximum occurs at an interior point c $\in$ (a, b). In what follows, we pick h small enough so that c + h always lies in (a, b). Then

If h > 0, $${\frac{{f(c+h) - f(c)}}{{h}}}$$$$\le$$0, and so $$\lim_{{h\to 0+}}^{}$$$${\frac{{f(c+h) - f(c)}}{{h}}}$$$$\le$$0, since we know the limit exists.

Similarly, if h < 0, $${\frac{{f(c+h) - f(c)}}{{h}}}$$$$\ge$$0, and so $$\lim_{{h\to 0+}}^{}$$$${\frac{{f(c+h) - f(c)}}{{h}}}$$$$\ge$$0, since we know the limit exists. Combining these, we see that f'(c) = 0, and we have the result in this case.

A similar argument applies if, instead, the global minimum occurs at the interior point c. The remaining situation occurs if both the global maximum and global minimum occur at end points; since f (a) = f (b), it follows that f is constant, and any c $\in$ (a, b) will do. $\qedsymbol$

Example 5.16   Investigate the number of roots of each of the polynomials

p(x) = x³ + 3x + 1    and    q(x) = x³ - 3x + 1.

Solution. Since p'(x) = 3(x² + 1) > 0 for all x $\in$ $\mathbb {R}$, we see that p has at most one root; for if it had two (or more) roots there would be a root of p'(x) = 0 between them by Rolle. Since p(0) = 1, while p(- 1) = - 3, there is at least one root by the Intermediate Value Theorem. Hence p has exactly one root.

We have q'(x) = 3(x² - 1) = 0 when x = ±1. Since q(- 1) = 3 and q(1) = - 1, there is a root of q between -1 and 1 by the Intermediate Value Theorem. Looking as x$\to$$\infty$ and as x$\to$ - $\infty$ shows here are three roots of q.

Exercise 5.17   Show that the equation x - e$\nolimits^{{-x}}_{}$ = 0 has exactly one root in the interval (0, 1).

Our version of Rolle's theorem is valuable as far as it goes, but the requirement that f (a) = f (b) is sufficiently strong that it can be quite hard to apply sometimes. Fortunately the geometrical description of the result -- that somewhere the tangent is parallel to the axis, does have a more general restatement.

Theorem 5.18 (The Mean Value Theorem)   Let f be continuous on [a, b], and differentiable on (a, b). Then there is some c with a < c < b such that

$${\frac{{f(b) - f(a)}}{{b-a}}}$$ = f'(c)    or equivalently    f (b) = f (a) + (b - a)f'(c).

Figure 5.2: Somewhere inside a chord, the tangent to f will be parallel to the chord. The accurate statement of this common-sense observation is the Mean Value Theorem.

Proof. We apply Rolle to a suitable function; let

h(x) = f (b) - f (x) - $${\frac{{f(b) - f(a)}}{{b-a}}}$$(b - x).

Then h is continuous on the interval [a, b], since f is, and in the same way, it is differentiable on the open interval (a, b). Also, f (b) = 0 and f (a) = 0. We can thus apply Rolle's theorem to h to deduce there is some point c with a < c < b such that h'(c) = 0. Thus we have

0 = h'(c) = - f'(c) + $${\frac{{f(b) - f(a)}}{{b-a}}}$$,

which is the required result. $\qedsymbol$

Example 5.19   The function f satisfies f'(x) = $${\frac{{1}}{{5-x^2}}}$$ and f (0) = 2. Use the Mean Value theorem to estimate f (1).

Solution. We first estimate the given derivative for values of x satisfying 0 < x < 1. Since for such x, we have 0 < x² < 1, and so 4 < 5 - x² < 5. Inverting we see that

$${\frac{{1}}{{5}}}$$ < f'(x) < $${\frac{{1}}{{4}}}$$    when    0 < x < 1.

Now apply the Mean Value theorem to f on the interval [0, 1] to obtain some c with 0 < c < 1 such that f (1) - f (0) = f'(c). From the given value of f (0), we see that 2.2 < f (1) < 2.25

Exercise 5.20   The function f satisfies f'(x) = $${\frac{{1}}{{5 + \sin x}}}$$ and f (0) = 0. Use the Mean Value theorem to estimate f ($\pi$/2).

Note the ``common sense'' description of what we have done. If the derivative doesn't change much, the function will behave linearly. Note also that this gives meaning to the approximation

f (a + h) $$\approx$$ f (a) + hf'(a).

We now see that the accurate version of this replaces f'(a) by f'(c) for some c between a and a + h.

Theorem 5.21   (The Cauchy Mean Value Theorem) Let f and g be both continuous on [a, b] and differentiable on (a, b). Then there is some point c with a < c < b such that

g'(c)$$\Big($$f (b) - f (a)$$\Big)$$ = f'(c)$$\Big($$g(b) - g(a)$$\Big)$$.

In particular, whenever the quotients make sense, we have

$${\frac{{f(b) - f(a)}}{{g(b)-g(a)}}}$$ = $${\frac{{f'(c)}}{{g'(c)}}}$$.

Proof. Let h(x) = f (x)$\Big($g(b) - g(a)$\Big)$ - g(x)$\Big($f (b) - f (a)$\Big)$, and apply Rolle's theorem exactly as we did for the Mean Value Theorem. Note first that since both f and g are continuous on [a, b], and differentiable on (a, b), it follows that h has these properties. Also h(a) = f (a)g(b) - g(a)f (b) = h(b). Thus we may apply Rolle to h, to deduce there is some point c with a < c < b such that h'(c) = 0. But

h'(c) = f'(c)$$\Big($$g(b) - g(a)$$\Big)$$ - g'(c)$$\Big($$f (b) - f (a)$$\Big)$$

Thus

f'(c)$$\Big($$g(b) - g(a)$$\Big)$$ = g'(c)$$\Big($$f (b) - f (a)$$\Big)$$

This is one form of the Cauchy Mean Value Theorem for f and g. If g'(c)$\ne$ 0 for any possible c, then the Mean Value theorem shows that g(b) - g(a)$\ne$ 0, and so we can divide the above result to get

$${\frac{{f(b) - f(a)}}{{g(b)-g(a)}}}$$ = $${\frac{{f'(c)}}{{g'(c)}}}$$,

giving a second form of the result. $\qedsymbol$

Note:Taking g(x) = x recovers the Mean Value Theorem.