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]
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
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
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.
To find the value or values of c that satisfy the equation, let's first calculate the necessary components of the equation.
Step 1: Find f(b) and f(a)
Given the function f(x) = x^(2/3), we need to find f(b) and f(a) using the given interval [0, 1].
For b = 1, f(b) = f(1) = 1^(2/3) = 1.
For a = 0, f(a) = f(0) = 0^(2/3) = 0.
Step 2: Find the derivative f'(c)
To calculate the derivative f'(c), we need to find the derivative of f(x) = x^(2/3).
Using the power rule, we get:
f'(x) = (2/3) * x^((2/3) - 1) = (2/3) * x^(-1/3) = 2x^(-1/3) / 3.
Step 3: Substitute the calculated values into the equation
We can now substitute the calculated values into the equation f(b) - f(a) / (b - a) = f'(c).
Substituting f(b) = 1, f(a) = 0, and f'(x) = 2x^(-1/3) / 3, we get:
1 - 0 / (1 - 0) = 2c^(-1/3) / 3.
Simplifying the equation, we have:
1 = 2c^(-1/3) / 3.
To solve for c, we can cross-multiply:
2c^(-1/3) = 3.
Now we can isolate c:
c^(-1/3) = 3/2.
To eliminate the exponent, we take the reciprocal of both sides:
c^(1/3) = 2/3.
To solve for c, we cube both sides:
c = (2/3)^3.
Simplifying the expression, we get:
c = 8/27.
Therefore, the value of c that satisfies the equation f(b)-f(a)/b-a = f'(c) for the given function f(x) = x^(2/3) and interval [0, 1] is c = 8/27.