Find the number c that satisfies the conclusion of the Mean Value Theorem.
f(x) = x/(x + 4)
[1, 8]
To apply the Mean Value Theorem to the given function f(x) = x/(x + 4) on the interval [1, 8], we need to verify two conditions:
1. f(x) is continuous on the closed interval [1, 8].
2. f(x) is differentiable on the open interval (1, 8).
First, let's check the continuity of f(x) on the closed interval [1, 8]. The function f(x) is a rational function, and since polynomials are continuous for all x, except where the denominator (x + 4) equals zero, we need to ensure that (x + 4) ≠ 0. Thus, the function f(x) is continuous on the interval [1, 8] as (x + 4) ≠ 0 for all x in [1, 8].
Next, we need to check the differentiability of f(x) on the open interval (1, 8). To do this, we will find the derivative of f(x) and check if it is defined on the open interval (1, 8).
The derivative of f(x) can be found using the quotient rule:
f'(x) = [(x + 4)(1) - (x)(1)] / (x + 4)^2
= (x + 4 - x) / (x + 4)^2
= 4 / (x + 4)^2
Since (x + 4)^2 is always positive, f'(x) exists for all x ≠ -4.
Now, we need to find the number c in the open interval (1, 8) that satisfies the conclusion of the Mean Value Theorem. According to the Mean Value Theorem, there exists a number c in (1, 8) such that f'(c) = [f(8) - f(1)] / (8 - 1).
Let's calculate f(8) and f(1) first:
f(8) = 8 / (8 + 4) = 8 / 12 = 2/3
f(1) = 1 / (1 + 4) = 1 / 5
So, [f(8) - f(1)] / (8 - 1) = (2/3 - 1/5) / 7 = (10/15 - 3/15) / 7 = 7/15 / 7 = 1/15
To find c, we need to solve the equation f'(c) = 1/15:
4 / (c + 4)^2 = 1/15
Multiply both sides by (c + 4)^2:
4 = (c + 4)^2 / 15
Multiply both sides by 15:
60 = (c + 4)^2
Taking the square root of both sides:
√60 = c + 4
Since we are looking for the value of c in the open interval (1, 8), we take the positive square root:
√60 ≈ 7.746
Finally, c ≈ √60 - 4 ≈ 7.746 - 4 ≈ 3.746
Therefore, the number c that satisfies the conclusion of the Mean Value Theorem is approximately 3.746.
To find the number c that satisfies the conclusion of the Mean Value Theorem for the function f(x) = x/(x + 4) on the interval [1, 8], we need to check two conditions:
1. f(x) must be continuous on the interval [a, b].
2. f(x) must be differentiable on the open interval (a, b).
Let's first check the continuity of f(x) on [1, 8]:
A function is continuous if it is defined and has no jumps, holes, or breaks in the function. In this case, f(x) = x/(x + 4) is defined and continuous for all real numbers except x = -4, where it is undefined. Since -4 does not lie within the interval [1, 8], f(x) is continuous on [1, 8].
Now let's check the differentiability of f(x) on (1, 8):
A function is differentiable if it has a derivative at every point in its domain. In this case, f(x) = x/(x + 4) is differentiable for all values of x except x = -4. Since -4 does not lie within the interval (1, 8), f(x) is differentiable on (1, 8).
Now we can apply the Mean Value Theorem:
The Mean Value Theorem states that if f(x) is continuous on [a, b] and differentiable on (a, b), then there exists at least one number c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
To find the value of c, we need to find the derivative of f(x) and evaluate it at some point in the interval (1, 8). Let's find the derivative of f(x):
f'(x) = [(x + 4) - x]/(x + 4)^2
= 4/(x + 4)^2
Now, let's find the values of f(8) and f(1):
f(8) = 8/(8 + 4)
= 8/12
= 2/3
f(1) = 1/(1 + 4)
= 1/5
Now, we can apply the Mean Value Theorem equation:
f'(c) = (f(8) - f(1))/(8 - 1)
= (2/3 - 1/5)/(7)
= (10/15 - 3/15)/7
= 7/15 / 7
= 1/15
We want to find the value of c such that f'(c) = 1/15. By solving the equation 4/(c + 4)^2 = 1/15, we can find c.
4/(c + 4)^2 = 1/15
Cross-multiplying:
15 * 4 = (c + 4)^2
60 = (c + 4)^2
Taking the square root of both sides:
±√60 = c + 4
Simplifying:
√60 = c + 4 or -√60 = c + 4
Subtracting 4 from both sides:
c = √60 - 4 or c = -√60 - 4
Therefore, the numbers c that satisfy the conclusion of the Mean Value Theorem for the function f(x) = x/(x + 4) on the interval [1, 8] are c = √60 - 4 and c = -√60 - 4.