Find the biggest value of c that satisfy the Mean Value Theorem for integrals for
f(x)= 1/(x+1)^6 on the interval [0,7]
the slope of the secant is (1/8^6 - 1)/7 ≈ -1/7
f' = -6/(x+1)^7
so, you want
-6/(c+1)^7 = -1/7
c ≈ 0.7
c is in the interval
Since f is monotonically decreasing, there is no other value for c.
To find the biggest value of c that satisfies the Mean Value Theorem for integrals, we need to find the derivative of the function f(x), and then determine the maximum value of the derivative on the given interval.
1. Start by finding the derivative of f(x). Let's call the derivative g(x):
g(x) = d/dx [f(x)] = d/dx [1/(x+1)^6]
To find the derivative, we can use the power rule and the chain rule:
1.1 Apply the power rule:
g(x) = -6/(x+1)^7.
2. Now, we need to find the maximum value of g(x) on the interval [0,7]. Since g(x) is a continuous function on a closed interval, the maximum value occurs either at the endpoints or at a critical point in the interval.
3. Check the endpoints of the interval:
g(0) = -6/(0+1)^7 = -6.
g(7) = -6/(7+1)^7 = -6/8^7.
4. Now, let's find the critical point(s) by setting g(x) equal to zero and solving for x:
-6/(x+1)^7 = 0.
Since the denominator (x+1)^7 cannot be zero, there are no critical points in the interval [0,7].
5. Now, compare the values at the endpoints and the critical points (if you have any).
g(0) = -6.
To compare g(0) with g(7), we need to simplify g(7):
g(7) = -6/8^7.
6. Now, compare the values:
g(0) = -6 < -6/8^7 = g(7).
7. Since g(7) is greater than g(0), g(7) is the maximum value of g(x) on the interval [0,7].
8. Therefore, the biggest value of c that satisfies the Mean Value Theorem for integrals is 7, where g(c) = g(7) = -6/8^7.
To find the biggest value of c that satisfies the Mean Value Theorem for integrals, we need to check the conditions of the Mean Value Theorem and find the maximum value of c.
The Mean Value Theorem for integrals 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 a value c in the open interval (a, b) such that the average value of f(x) over the interval [a, b] is equal to the value of the derivative of f(x) at c.
In this case, the function f(x) = 1/(x + 1)^6 is continuous on the closed interval [0, 7]. To check differentiability, we need to ensure that the function is defined and continuous on the open interval (0, 7).
Now, let's find the average value of f(x) over the interval [0, 7]. The average value is given by the formula:
Average value = 1/(b - a) * ∫[a, b] f(x) dx
Substituting the values a = 0 and b = 7, we have:
Average value = 1/(7 - 0) * ∫[0, 7] 1/(x + 1)^6 dx
To find the integral, we can use the power rule for integration:
∫(1/(x + a)^n) dx = -1/(n - 1)(x + a)^(n - 1)
Applying the power rule, we can find:
Average value = 1/7 * [-1/(6 - 1)(x + 1)^(6 - 1)] evaluated from 0 to 7
Simplifying, we get:
Average value = -1/7 *1/5 [(7 + 1)^5 - (0 + 1)^5]
Average value = -1/35 * (8^5 - 1)
Now, we know that the average value is equal to f'(c), where c is some value in the open interval (0, 7). Therefore, we have:
f'(c) = -1/35 * (8^5 - 1)
To find the maximum value of c, we need to find the maximum value of f'(c) within the given interval. Since f'(x) is a decreasing function when c > 0, we can find the maximum value by evaluating f'(x) at the endpoint x = 0.
f'(0) = -1/35 * (8^5 - 1)
Now, we can simplify f'(0) to find the largest value of c:
f'(0) = -1/35 * (32767)
f'(0) = -32767/35
Hence, the largest value of c that satisfies the Mean Value Theorem for integrals for f(x) = 1/(x + 1)^6 on the interval [0, 7] is -32767/35.