Verify the truth of Rolle's Theorem for the following functions :
`f(x)=(x-2)(x-4)^(2)` in the interval [2, 4].

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = (x - 2)(x - 4)^2 \) in the interval \([2, 4]\), we will follow these steps:

### Step 1: Check Continuity
We need to check if the function \( f(x) \) is continuous on the closed interval \([2, 4]\).

Since \( f(x) \) is a polynomial function, and all polynomial functions are continuous everywhere, it follows that \( f(x) \) is continuous on the interval \([2, 4]\).

### Step 2: Check Differentiability
Next, we need to check if \( f(x) \) is differentiable on the open interval \((2, 4)\).

Again, since \( f(x) \) is a polynomial function, it is differentiable everywhere, including the open interval \((2, 4)\).

### Step 3: Check the Values at the Endpoints
Now, we need to check the values of \( f(x) \) at the endpoints of the interval:

1. Calculate \( f(2) \):
\[ f(2) = (2 - 2)(2 - 4)^2 = 0 \cdot 4 = 0 \]

2. Calculate \( f(4) \):
\[ f(4) = (4 - 2)(4 - 4)^2 = 2 \cdot 0 = 0 \]

Since \( f(2) = f(4) = 0 \), the third condition of Rolle's Theorem is satisfied.

### Step 4: Find \( f'(x) \)
Now we will find the derivative \( f'(x) \) and check if there exists at least one \( c \) in the interval \((2, 4)\) such that \( f'(c) = 0 \).

Using the product rule:
\[ f'(x) = \frac{d}{dx}[(x - 2)(x - 4)^2] \]
Let \( u = (x - 2) \) and \( v = (x - 4)^2 \).

Using the product rule \( (uv)' = u'v + uv' \):
- \( u' = 1 \)
- \( v = (x - 4)^2 \) and \( v' = 2(x - 4) \)

Thus,
\[ f'(x) = (1)(x - 4)^2 + (x - 2)(2(x - 4)) \]
\[ = (x - 4)^2 + 2(x - 2)(x - 4) \]
\[ = (x - 4)^2 + 2(x^2 - 6x + 8) \]
\[ = (x - 4)^2 + 2x^2 - 12x + 16 \]

Now, we simplify \( f'(x) \):
\[ f'(x) = (x^2 - 8x + 16) + 2x^2 - 12x + 16 \]
\[ = 3x^2 - 20x + 32 \]

### Step 5: Solve \( f'(c) = 0 \)
Now we set \( f'(x) = 0 \):
\[ 3x^2 - 20x + 32 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3, b = -20, c = 32 \):
\[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot 32}}{2 \cdot 3} \]
\[ = \frac{20 \pm \sqrt{400 - 384}}{6} \]
\[ = \frac{20 \pm \sqrt{16}}{6} \]
\[ = \frac{20 \pm 4}{6} \]

Calculating the two possible values:
1. \( x = \frac{24}{6} = 4 \) (not in the open interval)
2. \( x = \frac{16}{6} = \frac{8}{3} \) (in the open interval)

### Conclusion
Since we found \( c = \frac{8}{3} \) which lies in the open interval \((2, 4)\), all conditions of Rolle's Theorem are satisfied.

Thus, we conclude that Rolle's Theorem is verified for the function \( f(x) = (x - 2)(x - 4)^2 \) in the interval \([2, 4]\).

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