Question

Determine if the function f(x)=44−x2‾‾‾‾‾‾√ satisfies the Mean Value Theorem on [0, 2]. If so, find all numbers c on the interval that satisfy the theorem.

Answer 1

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,

$f'(c) = \frac{f(b) - f(a)}{b-a}$

Here, the given function,

$f(x) = \sqrt{44-x^2}$

∵ 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],

$f'(x) = \frac{1}{2\sqrt{44-x^2}}(-2x) =-\frac{x}{\sqrt{44-x^2}}$

∵ f'(x) is defined for all values on the interval (0, 2),

Thus, f(x) is differentiable on (0, 2),

Now,

$f'(c) = -\frac{c}{\sqrt{44-c^2}}$

$\frac{f(2) - f(0)}{2-0}=\frac{\sqrt{44-4}-\sqrt{44}}{2}=\frac{\sqrt{40}-\sqrt{44}}{2}=\sqrt{10}-\sqrt{11}$

$-\frac{c}{\sqrt{44-c^2}}=\sqrt{10}-\sqrt{11}$

$\frac{c^2}{44-c^2}=10 + 11 - 2\sqrt{110}$

$c^2 = (44-c^2)(21-2\sqrt{110})$

$c^2 = 924 - 88\sqrt{110} - c^2(21 - 2\sqrt{110})$

$c^2 + (21 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}$

$(22 - 2\sqrt{110})c^2 = 924 - 88\sqrt{110}$

$c^2 = \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}$

$c=\sqrt{ \frac{924 - 88\sqrt{110}}{22 - 2\sqrt{110}}}\approx 1.012\in (0, 2)$

Hence, the function f(x) satisfies the Mean Value Theorem