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Math 1304

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.

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Math 1304

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  1. Math 1304 4.2 – The Mean Value Theorem

  2. 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].

  3. Rolle’s Theorem Theorem (Rolle’s Theorem) Let f be a function that satisfies the following three conditions • f is continuous on the closed interval [a,b] • f is differentiable on the open interval (a,b) • 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.

  4. The Mean Value Theorem Theorem (Mean ValueTheorem) Let f be a function that satisfies the following two conditions • f is continuous on the closed interval [a,b] • 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 conditions • g is continuous on the closed interval [a,b] • g is differentiable on the open interval (a,b) • 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).

  5. 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).

  6. 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).

  7. 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).

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