Verify mean value theorem for the function `f(x) = 1/(4x-1)` in `[1,4]`

Explanation

To verify the Mean Value Theorem (MVT) for the function \( f(x) = \frac{1}{4x - 1} \) on the interval \([1, 4]\), we will follow these steps:

### Step 1: Check the continuity of \( f(x) \) on \([1, 4]\)

The function \( f(x) = \frac{1}{4x - 1} \) is defined for all \( x \) except where the denominator is zero. We find the point where the denominator is zero:

\[ 4x - 1 = 0 \implies x = \frac{1}{4} \]

Since \( \frac{1}{4} \) is not in the interval \([1, 4]\), the function \( f(x) \) is continuous on \([1, 4]\).

### Step 2: Check the differentiability of \( f(x) \) on \((1, 4)\)

Next, we need to find the derivative \( f'(x) \). Using the quotient rule:

\[ f'(x) = \frac{d}{dx}\left(\frac{1}{4x - 1}\right) = \frac{0 \cdot (4x - 1) - 1 \cdot 4}{(4x - 1)^2} = \frac{-4}{(4x - 1)^2} \]

The derivative \( f'(x) \) exists for all \( x \) in the interval \((1, 4)\) since \( 4x - 1 \neq 0 \) for \( x \in (1, 4) \).

### Step 3: Apply the Mean Value Theorem

According to the Mean Value Theorem, there exists at least one \( c \in (1, 4) \) such that:

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

where \( a = 1 \) and \( b = 4 \).

First, we calculate \( f(1) \) and \( f(4) \):

\[ f(1) = \frac{1}{4(1) - 1} = \frac{1}{3} \]
\[ f(4) = \frac{1}{4(4) - 1} = \frac{1}{15} \]

Now, we can find \( f(b) - f(a) \):

\[ f(4) - f(1) = \frac{1}{15} - \frac{1}{3} = \frac{1}{15} - \frac{5}{15} = -\frac{4}{15} \]

Now, calculate \( b - a \):

\[ b - a = 4 - 1 = 3 \]

Now we can find the average rate of change:

\[ \frac{f(b) - f(a)}{b - a} = \frac{-\frac{4}{15}}{3} = -\frac{4}{45} \]

### Step 4: Set the derivative equal to the average rate of change

Now we set \( f'(c) \) equal to \(-\frac{4}{45}\):

\[ \frac{-4}{(4c - 1)^2} = -\frac{4}{45} \]

By simplifying, we can cancel \(-4\) from both sides:

\[ \frac{1}{(4c - 1)^2} = \frac{1}{45} \]

Taking the reciprocal gives:

\[ (4c - 1)^2 = 45 \]

Taking the square root on both sides:

\[ 4c - 1 = \pm \sqrt{45} = \pm 3\sqrt{5} \]

### Step 5: Solve for \( c \)

Now we solve for \( c \):

1. \( 4c - 1 = 3\sqrt{5} \)
\[ 4c = 3\sqrt{5} + 1 \implies c = \frac{3\sqrt{5} + 1}{4} \]

2. \( 4c - 1 = -3\sqrt{5} \)
\[ 4c = -3\sqrt{5} + 1 \implies c = \frac{-3\sqrt{5} + 1}{4} \]

### Step 6: Determine which \( c \) is in the interval \((1, 4)\)

Now we check which of these values is in the interval \((1, 4)\):

1. For \( c = \frac{3\sqrt{5} + 1}{4} \):
- Since \( \sqrt{5} \approx 2.236 \), \( 3\sqrt{5} \approx 6.708 \).
- So \( c \approx \frac{6.708 + 1}{4} \approx \frac{7.708}{4} \approx 1.927 \) (which is in \((1, 4)\)).

2. For \( c = \frac{-3\sqrt{5} + 1}{4} \):
- This value will be negative since \( -3\sqrt{5} \) is negative and thus not in the interval \((1, 4)\).

### Conclusion

Thus, we have verified the Mean Value Theorem for the function \( f(x) = \frac{1}{4x - 1} \) on the interval \([1, 4]\) with \( c = \frac{3\sqrt{5} + 1}{4} \).

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