Respuesta :
Answer:
x=1/3
Step-by-step explanation:
A function f is given as
[tex]f(x) = x^3-2x^2-4x+1[/tex] 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
[tex]f(1) = 1-2-4+1 = -4\\f(0) = 1[/tex]
The value
[tex]\frac{f(1)-f(0)}{1-0} =\frac{-4-1}{1} =-5[/tex]
Find derivative for f
[tex]f'(x) = 3x^2-4x-4[/tex]
Equate this to -5 to check mean value theorem
[tex]3x^2-4x-4=-5\\3x^2-4x+1=0\\\\(x-1)(3x-1) =0\\x= 1/3 : x = 1[/tex]
We find that 1/3 lies inside the interval (0,1)
