Suppose f(x) = x^3 on the interval [1, 4]. Use the Mean Value Theorem to find all values c in the open interval (1, 4) such that f'(c)= (f(4)-f(1))/4-1
c= square root of 7
c= cubed root of 21
c = 7
Mean Value Theorem does not apply
f(4) = 64
f(1) = 1
63/3 = 21 = average slope in interval
f' = 3 x^2 = 21
x = +/- sqrt 7
in interval
x = sqrt 7 = c
so A
To use the Mean Value Theorem to find the values of c that satisfy the equation f'(c) = (f(4) - f(1))/(4 - 1), we need to do the following steps:
1. Check if f(x) is continuous on the interval [1, 4].
2. Check if f(x) is differentiable on the open interval (1, 4).
3. If both conditions are met, find the derivative of f(x).
4. Set up the equation f'(c) = (f(4) - f(1))/(4 - 1).
5. Solve for c.
Now, let's go through each step:
1. f(x) = x^3 is a polynomial, and all polynomials are continuous everywhere. Therefore, f(x) is continuous on the interval [1, 4].
2. To check if f(x) is differentiable on the open interval (1, 4), we need to check if the derivative of f(x) exists for all x in (1, 4). Since f(x) = x^3 is a polynomial, it is differentiable for all real numbers. So, f(x) is differentiable on (1, 4).
3. The derivative of f(x) can be found by taking the derivative of each term:
f'(x) = 3x^2.
4. Set up the equation f'(c) = (f(4) - f(1))/(4 - 1):
3c^2 = (4^3 - 1^3)/(4 - 1).
5. Simplify the equation:
3c^2 = (64 - 1)/3.
3c^2 = 63/3.
3c^2 = 21.
c^2 = 21/3.
c^2 = 7.
c = ±√7.
So, the values of c that satisfy the equation f'(c) = (f(4) - f(1))/(4 - 1) are c = √7 and c = -√7.
However, since the interval given is (1, 4), only c = √7 lies within this interval. Therefore, the only value of c in the open interval (1, 4) that satisfies the given equation is c = √7.