Find a point c satisfying the conclusion of the Mean Value Theorem for the function f(x)= x^1/3 on the interval [1,8]

I got f'(c)= 1/7 but am not sure where to go from there.

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

f (1)=1^1/3=1
f (8) = 2

so, the point c must be such that
f'(c)= (f(8)-f(1))/(8-1)=(2-1)/7=1/7

but f'(c)=1/3 (c)^-2/3 and f'(c)=1/7 so
1/7=1/3(c^-2/3)
3/7=c^-2/3
take each side to the 3/2 power
c^2/3=7/3
c= cube root (7/3)^2 = 1.76

so, the conclusion of the theorem is borne out.

To find a point c that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(1/3) on the interval [1,8], we need to find a value that satisfies two conditions:

1. The function must be continuous on the interval [1,8].
2. The derivative of the function must be equal to the average rate of change of the function over the interval [1,8].

Let's break down the steps to find the point c:

Step 1: Verify that f(x) = x^(1/3) is continuous on the interval [1,8].
To check for continuity, we need to ensure that the function is defined and continuous at every point within the interval. Since f(x) = x^(1/3) is a polynomial function, it is defined and continuous for all real numbers, including the interval [1,8]. Therefore, f(x) = x^(1/3) is continuous on [1,8].

Step 2: Calculate the average rate of change of the function over the interval [1,8].
The average rate of change of a function over an interval [a,b] is given by the formula:
Average rate of change = (f(b) - f(a))/(b - a)

For our function f(x) = x^(1/3) on the interval [1,8], we have a = 1 and b = 8:
Average rate of change = (f(8) - f(1))/(8 - 1)
= (8^(1/3) - 1^(1/3))/7
= (2 - 1)/7
= 1/7

Step 3: Set f'(c) = (f(8) - f(1))/(8 - 1) = 1/7 and solve for c.
f'(x) represents the derivative of f(x). Since f'(c) equals the average rate of change of f(x) over the interval [1,8], we can set f'(c) = 1/7 and solve for c.

Taking the derivative of f(x) = x^(1/3), we get:
f'(x) = (1/3)x^(-2/3)

Setting f'(c) equal to 1/7, we have:
(1/3)c^(-2/3) = 1/7

To solve for c, we need to isolate the variable c:
c^(-2/3) = (1/7)(3/1)
c^(-2/3) = 3/7

Now, we solve for c by raising both sides to the power of -3/2:
(c^(-2/3))^(-3/2) = (3/7)^(-3/2)
c^1 = (7/3)^3
c = (7/3)^3

After evaluating the expression, we find that c β‰ˆ 3.792 is the point that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(1/3) on the interval [1,8], with f'(c) = 1/7.

To find a point c that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(1/3) on the interval [1, 8], you have correctly found that f'(c) = 1/7.

The Mean Value Theorem states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

In this case, we have f(x) = x^(1/3), and the interval [1, 8]. We know that a = 1 and b = 8.

Since we have f'(c) = 1/7, we can set this equal to (f(b) - f(a))/(b - a):

1/7 = (f(8) - f(1))/(8 - 1)

Now, let's calculate f(8) and f(1):

f(8) = 8^(1/3) = 2
f(1) = 1^(1/3) = 1

Substituting these values in, we have:

1/7 = (2 - 1)/(8 - 1)

Simplifying further, we get:

1/7 = 1/7

Since this equation is true, we have verified that f'(c) = 1/7 satisfies the conclusion of the Mean Value Theorem for the function f(x) = x^(1/3) on the interval [1, 8].