# 4.6 The Mean Value Theorem.

Calculus AB/BC – 5.1 Using the Mean Value Theorem
Calculus AB/BC – 5.1 Using the Mean Value Theorem

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4.6 The Mean Value Theorem

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Let f be a function that satisfies the following three hypotheses:
ROLLE’S THEOREM Let f be a function that satisfies the following three hypotheses: 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 number c in (a, b) such that f ‘(c) = 0.

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ROLLE’S THEOREM In each case, it appears there is at least one point (c, f(c)) on the graph where the tangent is horizontal and thus f ‘(c) = 0.

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ROLLE’S THEOREM Our main use of Rolle’s Theorem is in proving the following important theorem—which was first stated by another French mathematician, Joseph-Louis Lagrange. MEAN VALUE THEOREM

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Let f be a function that fulfills two hypotheses:
MEAN VALUE THEOREM Equations 1 and 2 Let f be a function that fulfills two hypotheses: f is continuous on the closed interval [a, b]. f is differentiable on the open interval (a, b). Then, there is a number c in (a, b) such that

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MEAN VALUE THEOREM The figures show the points A(a, f(a)) and B(b, f(b)) on the graphs of two differentiable functions.

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f ‘(c) is the slope of the tangent line at (c, f(c)).
MEAN VALUE THEOREM f ‘(c) is the slope of the tangent line at (c, f(c)). So, the Mean Value Theorem—in the form given by Equation 1—states that there is at least one point P(c, f(c)) on the graph where the slope of the tangent line is the same as the slope of the secant line AB.

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Proof: Let g(x) be the line going through the endpoints (a, f(a)) and (b, f(b)). The slope of g(x) = so g’(x) = for all x between a and b and g is continuous and differentiable. Let h(x) = f(x) – g(x) for [a, b]. h(x) satisfies the hypotheses of Rolle’s theorem: h(a) = f(a) – g(a) = 0 and h(b) = f(b) – g(b) = 0 h(x) is continuous for [a, b] since f and g are continuous for [a, b]. h(x) is differentiable for a < x < b since f and g are differentiable. So there is a value c in (a, b) such that h'(c) = 0. Since h(x) = f(x) – g(x) then h'(x) = f ‘(x) – g'(x) so we know that there is a number c, in (a, b), with h’(c) = 0. But 0 = h'(c) = f ‘(c) – g'(c). So f ‘(c) = g'(c) =

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MEAN VALUE THEOREM Example 1 To illustrate the Mean Value Theorem with a specific function, let’s consider f(x) = x3 – x, a = 0, b = 2. Since f is a polynomial, it is continuous and differentiable for all x. So, it is certainly continuous on [0, 2] and differentiable on (0, 2).

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MEAN VALUE THEOREM Example 3 Therefore, by the Mean Value Theorem, there is a number c in (0,2) such that:

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MEAN VALUE THEOREM Example 1

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The figure illustrates this calculation.
MEAN VALUE THEOREM Example 1 The figure illustrates this calculation. The tangent line at this value of c is parallel to the secant line OB.

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