If `f(x) = int_0^x t^2 (t - 2) (t -3) dt `for ` x in [0,oo]` then

Options

A. f has a local maximum at x= 2
B. f is decreasing on [2,3]
C. `EE` some ` c in ]0,oo` such that `f'( c) =0`
D. f has a local minimum at x = 3

Answer

To solve the problem, we need to find the local maxima and minima of the function defined by the integral:

\[ f(x) = \int_0^x t^2 (t - 2)(t - 3) dt \]

for \( x \) in the interval \([0, \infty)\).

### Step 1: Differentiate the function \( f(x) \)

Using the Fundamental Theorem of Calculus, we differentiate \( f(x) \) with respect to \( x \):

\[ f'(x) = \frac{d}{dx} \left( \int_0^x t^2 (t - 2)(t - 3) dt \right) = x^2 (x - 2)(x - 3) \]

### Step 2: Set the derivative equal to zero

To find the critical points, we set the derivative equal to zero:

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

### Step 3: Solve for \( x \)

The equation \( x^2 (x - 2)(x - 3) = 0 \) gives us the following solutions:

1. \( x^2 = 0 \) → \( x = 0 \)
2. \( x - 2 = 0 \) → \( x = 2 \)
3. \( x - 3 = 0 \) → \( x = 3 \)

Thus, the critical points are \( x = 0, 2, 3 \).

### Step 4: Determine the nature of the critical points

To determine whether these critical points are maxima or minima, we can use the first derivative test. We will analyze the sign of \( f'(x) \) around the critical points.

- For \( x < 0 \): \( f'(x) > 0 \) (since all factors are positive).
- For \( 0 < x < 2 \):
- Choose \( x = 1 \):
- \( f'(1) = 1^2(1 - 2)(1 - 3) = 1^2(-1)(-2) = 2 > 0 \) (increasing)
- For \( 2 < x < 3 \):
- Choose \( x = 2.5 \):
- \( f'(2.5) = (2.5)^2(2.5 - 2)(2.5 - 3) = (2.5)^2(0.5)(-0.5) < 0 \) (decreasing)
- For \( x > 3 \):
- Choose \( x = 4 \):
- \( f'(4) = (4)^2(4 - 2)(4 - 3) = 16(2)(1) > 0 \) (increasing)

### Step 5: Conclusion about maxima and minima

From the analysis:

- At \( x = 0 \): \( f'(x) \) changes from positive to positive (not a maximum or minimum).
- At \( x = 2 \): \( f'(x) \) changes from positive to negative (local maximum).
- At \( x = 3 \): \( f'(x) \) changes from negative to positive (local minimum).

### Final Results

- Local maximum at \( x = 2 \)
- Local minimum at \( x = 3 \)