To illustrate the Mean Value Theorem with a specific function, let\'s consider f

Question

To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = 5x5 - 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). therefore, by the Mean Value Theorem, there is a number c in (0, 2) such that

Now f(2) = ______, f(0) = ____, and f '(x) = _____, so this equation becomes

158= (________)2 = _____ c4 - ______

Which gives c^4= ______ that is, c =

To illustrate the Mean Value Theorem with a specific function, let's consider f(x) = 5x5 - 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). therefore, by the Mean Value Theorem, there is a number c in (0, 2) such that f(2) - f(0) = f '(c)(2 - 0) Now f(2) = , f(0) = , and f '(x) =, so this equation becomes Which gives c^4= that is, c = plusminus But c must be in (0, 2), so c The figure illustrates this calculation: the tangent line at this value of c is parallel to the secant line.

Explanation / Answer

Now f(2) = __158____, f(0) = _0___, and f '(x) = __25(x^4) - 1___, so this equation becomes

158= (_25(c^4) - 1_______)2 = ___50__ c4 - ___2___

Which gives c^4= _3.2_____ that is, c =

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