calculus

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Find the value or values of c that satisfy the equation
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 -

y = 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 -

y = 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 -

The answer in my textbook says 1 for the value of c. I don't understand how it arrived there.

• calculus -

Nor do I

• calculus -

Sounds like they wanted f'(c) rather than c.