Respuesta :
Answer:
The function f(x) satisfies the Mean Value Theorem
Step-by-step explanation:
Mean Value Theorem states that if f be a function such that,
- It is continuous on [a, b].
- It is differentiable on (a, b).
Then there is at least a number c in (a, b) such that,
[tex]f'(c) = \frac{f(b) - f(a)}{b-a}[/tex]
Here, the given function,
[tex]f(x) = \sqrt{44-x^2}[/tex]
∵ f(x) is defined for all real values x for which 44-x² ≥ 0
44 ≥ x² ⇒ ±√44 ≥ x ⇒-√44 ≤ x ≤ √44
Thus, f(x) is continuous on [0, 2],
[tex]f'(x) = \frac{1}{2\sqrt{44-x^2}}(-2x) =-\frac{x}{\sqrt{44-x^2}}[/tex]
∵ f'(x) is defined for all values on the interval (0, 2),
Thus, f(x) is differentiable on (0, 2),
Now,
[tex]f'(c) = -\frac{c}{\sqrt{44-c^2}}[/tex]
[tex]\frac{f(2) - f(0)}{2-0}=\frac{\sqrt{44-4}-\sqrt{44}}{2}=\frac{\sqrt{40}-\sqrt{44}}{2}=\sqrt{10}-\sqrt{11}[/tex]
[tex]-\frac{c}{\sqrt{44-c^2}}=\sqrt{10}-\sqrt{11}[/tex]
[tex]\frac{c^2}{44-c^2}=10 + 11 - 2\sqrt{110}[/tex]
[tex]c^2 = (44-c^2)(21-2\sqrt{110})[/tex]
[tex]c^2 = 924 - 88\sqrt{110} - c^2(21 - 2\sqrt{110})[/tex]
[tex]c^2 + (21 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}[/tex]
[tex](22 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}[/tex]
[tex]c^2 = \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}[/tex]
[tex]c=\sqrt{ \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}}\approx 1.012\in (0, 2)[/tex]
Hence, the function f(x) satisfies the Mean Value Theorem
The number c on the interval that satisfies the mean value theorem is; c = 1.0118 and thus the function satisfies the mean value theorem
How to prove the mean value theorem?
Mean Value Theorem states that if f is a function such that;
It is continuous on [a, b] and It is differentiable on (a, b).
Then, there is at least a number c in (a, b) such that;
f'(c) = [f(b) - f(a)]/(b - a)
We have the function;
f(x) = √(44 - x²)
where;
f(x) is defined for all real values x for which 44 - x² ≥ 0
Thus;
44 ≥ x²
⇒ ±√44 ≥ x
Thus, -√44 ≤ x ≤ √44
From the question, f(x) is continuous on [0, 2].
f'(x) = -x/√(44 - x²)
Then by mean value theorem;
f'(c) = -c/√(44 - c²)
Then at 0,2;
f'(c) = [f(2) - f(0)]/(2 - 0)
f'(c) = √10 - √11
Thus;
-c/√(44 - c²) = √10 - √11
Simplifying this to solve for c on an online calculator gives;
c = 1.0118
This means c falls within the giving range of [0, 2] and as such the function satisfies the mean value theorem.
Read more about Mean Value Theorem at; https://brainly.com/question/19052862
