Question

Determine whether the Mean Value Theorem can f(x)=(x+8)ln(x+8),[-7,-6]

Answer 1

The Mean Value Theorem is applicable to the function f(x) = (x+8)ln(x+8) on the interval [-7, -6].

Step-by-step explanation:

The Mean Value Theorem (MVT) states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that the instantaneous rate of change (the derivative) at c is equal to the average rate of change over the interval [a, b]. Mathematically, this can be expressed as f'(c) = [f(b) - f(a)] / (b - a).

In this case, the function f(x) = (x+8)ln(x+8) is continuous and differentiable on the interval [-7, -6]. To apply the MVT, we first find the derivative of f(x). Using the product rule and the chain rule, we get f'(x) = ln(x+8) + 1.

Now, we can calculate the average rate of change over the given interval: [f(-6) - f(-7)] / (-6 - (-7)). Substituting the values, we get [(-6+8)ln(-6+8) - (-7+8)ln(-7+8)] / 1.

Simplifying further, we have [2ln2 + ln1] / 1. Since ln1 is zero, the expression becomes 2ln2. Therefore, the average rate of change is 2ln2.

Now, we set the derivative equal to this average rate of change and solve for c: ln(c+8) + 1 = 2ln2. Subtracting 1 and rearranging, we get ln(c+8) = 2ln2 - 1. Exponentiating both sides, we find c+8 =
`e^(^2^l^n^2^-^1^)`, and solving for c, we get c =
`e^(^2^l^n^2^-^1^)` - 8. This value of c satisfies the conditions of the Mean Value Theorem for the given function and interval.