Question

f(x) = 4-(6-x)^{2/3} in [5, 7]

• A. Lagranges theorem is applicable
• B. Rolle's theorem is applicable
• C. Lagranges theorem is applicable but not the Rolle's theorem
• D. Both theorems are not applicable

Answer 1

Answer: (D)

D) Both theorems are not applicable

Answer 2

To determine the applicability of Rolle's Theorem and Lagrange's Mean Value Theorem (LMVT) for the function $f(x) = 4-(6-x)^{2/3}$ in the interval $[5, 7]$, we need to check their conditions.

Conditions for Rolle's Theorem:

1. $f(x)$ is continuous on the closed interval $[a, b]$.
2. $f(x)$ is differentiable on the open interval $(a, b)$.
3. $f(a) = f(b)$.

Conditions for Lagrange's Mean Value Theorem (LMVT):

1. $f(x)$ is continuous on the closed interval $[a, b]$.
2. $f(x)$ is differentiable on the open interval $(a, b)$.

Let's check these conditions for $f(x) = 4-(6-x)^{2/3}$ in $[5, 7]$.

1. Continuity:

The function $g(u) = u^{2/3}$ is continuous for all real numbers $u$. The function $h(x) = 6-x$ is a polynomial, and thus continuous for all real numbers $x$. Since $f(x) = 4 - (h(x))^{2/3}$ is a composition of continuous functions and a constant, it is continuous for all real numbers $x$. Therefore, $f(x)$ is continuous on the closed interval $[5, 7]$. Condition 1 for both theorems is satisfied.

2. Differentiability:

We need to find the derivative $f'(x)$:

$f'(x) = \frac{d}{dx} \left[4 - (6-x)^{2/3}\right]$

Using the chain rule, $\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}$, where $u = (6-x)$ and $n = 2/3$.

$f'(x) = 0 - \frac{2}{3}(6-x)^{\frac{2}{3}-1} \cdot \frac{d}{dx}(6-x)$

$f'(x) = - \frac{2}{3}(6-x)^{-1/3} \cdot (-1)$

$f'(x) = \frac{2}{3}(6-x)^{-1/3}$

$f'(x) = \frac{2}{3\sqrt[3]{6-x}}$

For $f(x)$ to be differentiable on the open interval $(5, 7)$, $f'(x)$ must exist for all $x \in (5, 7)$. The derivative $f'(x)$ is undefined when the denominator is zero, i.e., $3\sqrt[3]{6-x} = 0$. This occurs when $6-x = 0$, which means $x = 6$. The point $x = 6$ lies within the open interval $(5, 7)$. Since $f'(x)$ is not defined at $x=6$, the function $f(x)$ is not differentiable at $x=6$. Therefore, $f(x)$ is not differentiable on the open interval $(5, 7)$. Condition 2 for both theorems is NOT satisfied.

Since the differentiability condition is not met, neither Rolle's Theorem nor Lagrange's Mean Value Theorem is applicable.

3. Check $f(a) = f(b)$ for Rolle's Theorem (optional, as differentiability fails):

$f(5) = 4 - (6-5)^{2/3} = 4 - (1)^{2/3} = 4 - 1 = 3$.
$f(7) = 4 - (6-7)^{2/3} = 4 - (-1)^{2/3} = 4 - ((-1)^2)^{1/3} = 4 - (1)^{1/3} = 4 - 1 = 3$.
So, $f(5) = f(7)$. Even though this condition is met, Rolle's Theorem is not applicable because the differentiability condition failed.

Conclusion:

Since the function $f(x)$ is not differentiable on the open interval $(5, 7)$ (specifically at $x=6$), neither Rolle's Theorem nor Lagrange's Mean Value Theorem is applicable.