f (x) = x^2, [4, 5]
If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that
f '(c) = f (b) − f (a)/b − a
answers as a comma-separated list.
well, what's the problem?
f(5)-f(4) = 25-16 = 9
5-4 = 1
f' = 2x
so, where does 2x = 9?
4.5 or 9/2 thnks for some reason I forgot to put 2x I kept doing x=9! thank you :)
To find all values of c in the open interval (a, b) where the Mean Value Theorem can be applied, we need to check if the function f(x) = x^2 satisfies the conditions of the Mean Value Theorem on the interval [4, 5].
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 number c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a)
Let's verify the conditions of the Mean Value Theorem for the function f(x) = x^2 on the interval [4, 5].
1. Continuity on the closed interval [4, 5]:
The function f(x) = x^2 is a polynomial, and all polynomials are continuous for all real numbers. Thus, f(x) = x^2 is continuous on the closed interval [4, 5].
2. Differentiability on the open interval (4, 5):
The derivative of f(x) = x^2 is f'(x) = 2x, which is also a polynomial and thus differentiable for all real numbers. Therefore, f(x) = x^2 is differentiable on the open interval (4, 5).
Since f(x) = x^2 satisfies the conditions of the Mean Value Theorem on the interval [4, 5], we can proceed to find the values of c.
Now, let's calculate:
f'(x) = 2x,
f'(4) = 2*4 = 8,
f'(5) = 2*5 = 10,
To find the values of c, we need to solve the equation f'(c) = (f(b) - f(a))/(b - a), which is:
2c = (f(5) - f(4))/(5 - 4),
2c = (25 - 16)/1,
2c = 9,
c = 9/2.
So, the values of c in the open interval (4, 5) that satisfy the Mean Value Theorem for f(x) = x^2 are: c = 9/2.
Therefore, the answer is c = 9/2.