2. Determine if the Mean Value Theorem can be applied theorem . If it cannot, explain why not. (i) \( y=(x-3)^{\frac{2}{3}},[1,4] \) (ii) \( y=-(-5 x+25)^{\frac{1}{2}},[3,5] \)

To determine whether the **Mean Value Theorem (MVT)** can be applied to each function on the given intervals, we need to verify two key conditions: 1. **Continuity** on the closed interval \([a, b]\). 2. **Differentiability** on the open interval \((a, b)\). The **Mean Value Theorem** states that if a function satisfies these two conditions, then there exists at least one point \(c \in (a, b)\) where the instantaneous rate of change (derivative) equals the average rate of change over \([a, b]\). Let's analyze each function: --- ### (i) \( y = (x - 3)^{\frac{2}{3}} \) on \([1, 4]\) 1. **Continuity:** - The function \( y = (x - 3)^{\frac{2}{3}} \) is continuous for all real numbers \(x\) because it is a composition of continuous functions (polynomial and root functions). 2. **Differentiability:** - Compute the derivative: \[ y' = \frac{2}{3}(x - 3)^{-\frac{1}{3}} \] - The derivative \( y' \) is undefined at \( x = 3 \) because it involves division by zero (\( (x - 3)^{-\frac{1}{3}} \) becomes infinite as \( x \) approaches 3). - Since \( x = 3 \) lies within the open interval \((1, 4)\), the function is **not differentiable** at this point. 3. **Conclusion:** - **MVT cannot be applied** because the function fails to be differentiable on the entire open interval \((1, 4)\). --- ### (ii) \( y = -\sqrt{-5x + 25} \) on \([3, 5]\) 1. **Continuity:** - The expression inside the square root, \(-5x + 25\), must be non-negative: \[ -5x + 25 \geq 0 \Rightarrow x \leq 5 \] - On the interval \([3, 5]\), this condition is satisfied (since \(x \leq 5\)), ensuring that \( y \) is real and continuous. 2. **Differentiability:** - Compute the derivative: \[ y' = \frac{d}{dx} \left( -\sqrt{-5x + 25} \right) = -\frac{1}{2}(-5x + 25)^{-\frac{1}{2}} \cdot (-5) = \frac{5}{2\sqrt{-5x + 25}} \] - The derivative \( y' \) is defined for all \( x \in (3, 5) \) because \(-5x + 25 > 0\) in this open interval, ensuring that the denominator never becomes zero. - Thus, \( y \) is **differentiable** on the open interval \((3, 5)\). 3. **Conclusion:** - **MVT can be applied** because the function is both continuous on \([3, 5]\) and differentiable on \((3, 5)\). --- ### Summary - **(i)** The Mean Value Theorem **cannot** be applied to \( y = (x - 3)^{\frac{2}{3}} \) on \([1, 4]\) because the function is not differentiable at \( x = 3 \), which lies within the interval. - **(ii)** The Mean Value Theorem **can** be applied to \( y = -\sqrt{-5x + 25} \) on \([3, 5]\) since the function is continuous on \([3, 5]\) and differentiable on \((3, 5)\). Answer: Problem (i) Answer: The Mean Value Theorem cannot be applied to \( y = (x - 3)^{2/3} \) on [1, 4] because the function is not differentiable at \( x = 3 \), which lies within the interval. Problem (ii) Answer: The Mean Value Theorem can be applied to \( y = -\sqrt{-5x + 25} \) on [3, 5] since the function is continuous on [3, 5] and differentiable on (3, 5).