The points of extemum of `f(x)= int_(0)^(x^(2)) (t^(2)- 5t +4)/(2+e^(t)) dt` are

Options

A. x=1
B. `x= -1`
C. x=0
D. `x= -2`

Answer

To find the points of extremum of the function

\[ f(x) = \int_{0}^{x^{2}} \frac{t^{2} - 5t + 4}{2 + e^{t}} \, dt, \]

we need to follow these steps:

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

To find the points of extremum, we first need to differentiate \( f(x) \) with respect to \( x \). We will apply the Fundamental Theorem of Calculus and the chain rule.

\[ f'(x) = \frac{d}{dx} \left( \int_{0}^{x^{2}} \frac{t^{2} - 5t + 4}{2 + e^{t}} \, dt \right) = \frac{(x^{2})^{2} - 5(x^{2}) + 4}{2 + e^{x^{2}}} \cdot \frac{d}{dx}(x^{2}) \]

Using the chain rule, we have:

\[ f'(x) = \frac{x^{4} - 5x^{2} + 4}{2 + e^{x^{2}}} \cdot 2x \]

### Step 2: Set the derivative to zero

To find the points of extremum, we set \( f'(x) = 0 \):

\[ \frac{x^{4} - 5x^{2} + 4}{2 + e^{x^{2}}} \cdot 2x = 0 \]

This equation will be zero if either \( 2x = 0 \) or the numerator is zero.

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

1. From \( 2x = 0 \):
\[ x = 0 \]

2. From the numerator \( x^{4} - 5x^{2} + 4 = 0 \):
Let \( y = x^{2} \), then we have:
\[ y^{2} - 5y + 4 = 0 \]
This is a quadratic equation. We can use the quadratic formula:
\[ y = \frac{5 \pm \sqrt{(-5)^{2} - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \]
Thus, we get:
\[ y = \frac{8}{2} = 4 \quad \text{or} \quad y = \frac{2}{2} = 1 \]

Therefore, \( x^{2} = 4 \) gives \( x = \pm 2 \) and \( x^{2} = 1 \) gives \( x = \pm 1 \).

### Step 4: List all points of extremum

Combining all the solutions, we have the points of extremum:
\[ x = 0, \quad x = 1, \quad x = -1, \quad x = 2, \quad x = -2 \]

### Final Answer

The points of extremum of the function \( f(x) \) are \( x = 0, \, x = 1, \, x = -1, \, x = 2, \, x = -2 \).

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