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
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]
5 answers
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
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
The answer in my textbook says 1 for the value of c. I don't understand how it arrived there.
Nor do I
Sounds like they wanted f'(c) rather than c.
1 answer