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At what values of x does f(x) = x^3 - 2x^2 -4x+1 satisfy the mean value theorwm on [0,1]

User Zachbugay
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1 Answer

4 votes

Answer:

x=1/3

Explanation:

A function f is given as


f(x) = x^3-2x^2-4x+1 in the interval [0,1]

This function f being an algebraic polynomial is continuous in the interval [0,1] and also f is differntiable in the open interval (0,1)

Hence mean value theorem applies for f in the given interval


f(1) = 1-2-4+1 = -4\\f(0) = 1

The value


(f(1)-f(0))/(1-0) =(-4-1)/(1) =-5

Find derivative for f


f'(x) = 3x^2-4x-4

Equate this to -5 to check mean value theorem


3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1

We find that 1/3 lies inside the interval (0,1)

User Sophie Crommelinck
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