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January 31, 2015

January 31, 2015

Posted by **Aubrey** on Tuesday, July 22, 2014 at 7:00pm.

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

in the conclusion of the mean value theorem for the given function and interval

f(x)= x^(2/3) , [0,1]

- calculus -
**Damon**, Tuesday, July 22, 2014 at 7:22pmy = x^(2/3)

at x = 1 which is b

y = 1^(2/3) = 1

at x = 0 which is a

y = 1^0 = 1

y(1) - y(0) = 1 - 0 = 1

b - a = 1 - 0 = 1

so

1/1 = 1

so according to the mean value theorem somewhere between x = 0 and x = 1 the derivative = 1

dy/dx = (2/3) x^-1/3 = (2/3)/x^(1/3)

when is that equal to one?

x^(1/3) = 2/3

(1/3) log x = log 2/3

log x = -.52827

x= .296

- calculus typo -
**Damon**, Tuesday, July 22, 2014 at 7:24pmy = x^(2/3)

at x = 1 which is b

y = 1^(2/3) = 1

at x = 0 which is a

y = 0^0 = 0

y(1) - y(0) = 1 - 0 = 1

b - a = 1 - 0 = 1

so

1/1 = 1

so according to the mean value theorem somewhere between x = 0 and x = 1 the derivative = 1

dy/dx = (2/3) x^-1/3 = (2/3)/x^(1/3)

when is that equal to one?

x^(1/3) = 2/3

(1/3) log x = log 2/3

log x = -.52827

x= .296

- calculus -
**Aubrey**, Tuesday, July 22, 2014 at 7:26pmThe answer in my textbook says 1 for the value of c. I don't understand how it arrived there.

- calculus -
**Damon**, Tuesday, July 22, 2014 at 7:35pmNor do I

- calculus -
**Steve**, Tuesday, July 22, 2014 at 7:37pmSounds like they wanted f'(c) rather than c.

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