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Theorems in Calculus: Extreme Value Theorem, Rolle's Theorem, and Mean Value Theorem

Learn about the Extreme Value Theorem, Rolle's Theorem, and the Mean Value Theorem in calculus, including proofs and applications. Understand how these theorems relate to functions and derivatives.

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Theorems in Calculus: Extreme Value Theorem, Rolle's Theorem, and Mean Value Theorem

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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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