Math 1304

4.2 – The Mean Value Theorem

Recall The Extreme Value Theorem

• Theorem (Extreme Value Theorem) Let f be a continuous function of the closed interval [a,b]. Then f obtains an absolute maximum and an absolute minimum on the interval [a,b].

Rolle’s Theorem

Theorem (Rolle’s Theorem) Let f be a function that satisfies the following three conditions

1) f is continuous on the closed interval [a,b]2) f is differentiable on the open interval (a,b)3) f(a) = f(b)

Then there is a real number c in the open interval (a,b) such that f’(c)=0.

Proof: We will use the Extreme Value Theorem.There are three cases: a) f is constant, b) f has a value f(c’) > f(a) for

some c in (a,b), c) f has a value f(c’) < f(a) for some c in (a,b). In case a) f’(c) = 0 for all c in (a,b).In case b) f has a maximum at some c in (a,b) by the Extreme Value

Theorem. By Fermat’s Theorem f’(c)=0.In case c) f has a minimum at some c in (a,b) by the Extreme Value

Theorem. By Fermat’s Theorem f’(c)=0. Hence f’(c)=0.

The Mean Value TheoremTheorem (Mean ValueTheorem) Let f be a function that satisfies the following two

conditions1) f is continuous on the closed interval [a,b]2) f is differentiable on the open interval (a,b)

Then there is a real number c in the open interval (a,b) such that f’(c)=(f(b)-f(a))/(b-a).

Proof: We will use Rolle’s Theorem on the function g(x) = f(x) – f(a) - (f(b)-f(a))/(b-a) (x – a).Note that

g(a) = f(a) – f(a) - (f(b)-f(a))/(b-a) (a-a) = 0

g(b) = f(b) – f(a) - (f(b)-f(a))/(b-a) (b-a) = f(b) – f(a) – (f(b) – f(a) = 0

g’(x) = f’(x) - (f(b)-f(a))/(b-a)

The function g satisfies the conditions1) g is continuous on the closed interval [a,b]2) g is differentiable on the open interval (a,b)3) g(a) = 0 = g(b)

Thus g’(c)=0 for some c in (a,b). Thus f’(c) - (f(b)-f(a))/(b-a) = 0 for this point c.Thus f’(c) = (f(b)-f(a))/(b-a) for this point c in the open interval (a,b).

Zero Derivative

Theorem: If f’(x) is zero for all x in an interval (a,b), then f is constant

Proof:

Pick any two points x1 and x2 in (a,b) with x1<x2.

Then x2-x1 > 0.

By the Mean Value Theorem, there is a point c in (x1,x2) such that f’(c)=(f(x2)-f(x1))/(x2-x1).

Thus (f(x2)-f(x1))/(x2-x1) = 0.

Thus f(x2)-f(x1)=0.

Thus f(x2) = f(x1).

Positive Derivative

Theorem: If f’(x) is positive for all x in an interval (a,b), then f is increasing

Proof:

Pick any two points x1 and x2 in (a,b) with x1<x2.

Then x2-x1 > 0.

By the Mean Value Theorem, there is a point c in (x1,x2) such that f’(c)=(f(x2)-f(x1))/(x2-x1).

Thus (f(x2)-f(x1))/(x2-x1) > 0.

Thus f(x2)-f(x1) > 0.

Thus f(x2) > f(x1).

Negative Derivative

Theorem: If f’(x) is negative for all x in an interval (a,b), then f is decreasing.

Proof:

Pick any two points x1 and x2 in (a,b) with x1<x2.

Then x2-x1 > 0.

By the Mean Value Theorem, there is a point c in (x1,x2) such that f’(c)=(f(x2)-f(x1))/(x2-x1).

Thus (f(x2)-f(x1))/(x2-x1) < 0.

Thus f(x2)-f(x1) < 0.

Thus f(x2) < f(x1).