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Morgarella [4.7K]
3 years ago
6

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

Mathematics
1 answer:
denis-greek [22]3 years ago
4 0

Answer:

x=1/3

Step-by-step 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

\frac{f(1)-f(0)}{1-0} =\frac{-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)

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