Find all values of c that satisfy the Mean Value Theorem for f(x) = x^3 + 1 on [2, 4].
just find c such that
3c^2 = (65-9)/(4-2)
To determine the values of c that satisfy the Mean Value Theorem for the function f(x) on an interval [a, b], we need to check two conditions:
1. f(x) must be continuous on the closed interval [a, b].
2. f(x) must be differentiable on the open interval (a, b).
Let's apply these conditions to the given function f(x) = x^3 + 1 on the interval [2, 4].
1. Continuity: To check if f(x) is continuous on [2, 4], we need to ensure that there are no holes, asymptotes, or jumps in the graph of f(x) within that interval. The function f(x) = x^3 + 1 is a polynomial function, and polynomial functions are continuous for all real numbers. So, f(x) = x^3 + 1 is continuous on [2, 4].
2. Differentiability: To check if f(x) is differentiable on (2, 4), we need to ensure that the derivative of f(x) exists for all values of x in the open interval (2, 4). The derivative of f(x) = x^3 + 1 is f'(x) = 3x^2. The derivative exists for all real numbers, so it exists for all values of x in the open interval (2, 4).
Since both conditions are satisfied, we can conclude that the Mean Value Theorem applies to the function f(x) = x^3 + 1 on the interval [2, 4]. This means that there exists at least one value c in the open interval (2, 4) such that the derivative of f(x) at c is equal to the average rate of change of f(x) on the interval [2, 4]:
f'(c) = (f(4) - f(2)) / (4 - 2)
To find the values of c, we first calculate the average rate of change:
f'(c) = (f(4) - f(2)) / (4 - 2)
f'(c) = (4^3 + 1 - 2^3 - 1) / 2
f'(c) = (64 + 1 - 8 - 1) / 2
f'(c) = 56 / 2
f'(c) = 28
Now, we need to find the value of c such that f'(c) = 28. Taking the derivative of f(x) = x^3 + 1 gives us f'(x) = 3x^2. Setting this equal to 28 and solving for x:
3x^2 = 28
x^2 = 28 / 3
x = ±√(28 / 3)
Therefore, there are two values of c that satisfy the Mean Value Theorem for f(x) = x^3 + 1 on [2, 4], which are approximately x ≈ ±1.964.