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f(x)=x^3 -x on (0,2)

Calculus
Tutor: None Selected Time limit: 1 Day

show that the mean value theorem applies and find the value of c guaranteed by the theorem

Apr 21st, 2015

a = 0   b = 2

f(x) = x^3 - x

f '(x) = 3x^2 - 1

f(a) = (a)^3 - (a) = (0)^3 - (0) = 0

f(b) = (b)^3 - (b) = (2)^3 - (2) = 8 - 2 = 6

Now the formula for mean value is

f '(x) = f(b) - f(a) / b - a

f '(x) = 6 - 0 / 2 - 0

f '(x) = 6 / 2

f '(x) = 3

Now to find the value of c

Put x = c in f'(x) = 3x^2 - 1  , then

f'(c) = 3c^2 - 1

Now

3c^2 - 1 = 3

3c^2 = 3 + 1

3c^2 = 4

c^2 = 4/3

c = +√4/3       or      c = -√4/3

Apr 21st, 2015

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