The function
`f(x) = int_1^x { 2 (t -1) (t - 2)^3 + 3 (t -1)^2 (t - 2)^2} dt` attains its maximum at x =

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the maximum of the function

\[ f(x) = \int_1^x \left( 2(t - 1)(t - 2)^3 + 3(t - 1)^2(t - 2)^2 \right) dt, \]

we will follow these steps:

### Step 1: Differentiate the Function
To find the critical points where the function may attain a maximum, we first need to differentiate \( f(x) \) with respect to \( x \). By the Fundamental Theorem of Calculus, we have:

\[ f'(x) = 2(x - 1)(x - 2)^3 + 3(x - 1)^2(x - 2)^2. \]

### Step 2: Set the Derivative to Zero
Next, we set the derivative equal to zero to find the critical points:

\[ f'(x) = 0. \]

This gives us:

\[ 2(x - 1)(x - 2)^3 + 3(x - 1)^2(x - 2)^2 = 0. \]

### Step 3: Factor the Derivative
We can factor out the common terms:

\[ (x - 1)(x - 2)^2 \left( 2(x - 2) + 3(x - 1) \right) = 0. \]

This simplifies to:

\[ (x - 1)(x - 2)^2 (5x - 7) = 0. \]

### Step 4: Solve for Critical Points
Setting each factor to zero gives us the critical points:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( (x - 2)^2 = 0 \) → \( x = 2 \)
3. \( 5x - 7 = 0 \) → \( x = \frac{7}{5} \)

### Step 5: Analyze the Critical Points
We have three critical points: \( x = 1, x = 2, \) and \( x = \frac{7}{5} \). We will use the number line method to determine where \( f'(x) \) changes sign.

### Step 6: Test Intervals
We will test the sign of \( f'(x) \) in the intervals created by the critical points:

- For \( x < 1 \): Choose \( x = 0 \)
- \( f'(0) < 0 \) (negative)

- For \( 1 < x < \frac{7}{5} \): Choose \( x = 1.4 \)
- \( f'(1.4) > 0 \) (positive)

- For \( \frac{7}{5} < x < 2 \): Choose \( x = 1.6 \)
- \( f'(1.6) < 0 \) (negative)

- For \( x > 2 \): Choose \( x = 3 \)
- \( f'(3) > 0 \) (positive)

### Step 7: Determine Maximum
From the sign analysis, we see that \( f'(x) \) changes from negative to positive at \( x = 1 \) and from positive to negative at \( x = \frac{7}{5} \). Thus, the function \( f(x) \) attains its maximum at:

\[ x = \frac{7}{5}. \]

### Conclusion
The function \( f(x) \) attains its maximum at \( x = \frac{7}{5} \).

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