Question

The function \(f(x)=xe^{x(1-x)}\),x∈R is

• A. increasing in \((-\frac{1}{2} , 1)\)
• B. increasing in \((-\frac{1}{2} , 1)\)
• C. increasing in  \((-1 , -\frac{1}{2})\)
• D. decreasing in \((-\frac{1}{2} , \frac{1}{2})\)

Answer 1

The Correct Option is A

To determine the intervals where the function \( f(x) = xe^{x(1-x)} \) is increasing or decreasing, we need to analyze the first derivative of the function.

Step 1: Find the first derivative of \( f(x) \).

The function is \( f(x) = xe^{x(1-x)} \). We use the product rule for differentiation, which states that if \( u(x) = x \) and \( v(x) = e^{x(1-x)} \), then:

\( f'(x) = u'(x)v(x) + u(x)v'(x) \)

Where \( u'(x) = 1 \) and for \( v(x) = e^{x(1-x)} \), we use the chain rule to find \( v'(x) \).

Step 2: Differentiate \( v(x) = e^{x(1-x)} \).

The derivative of \( v(x) \) is:

\( v'(x) = e^{x(1-x)} \cdot \frac{d}{dx}(x(1-x)) \)

Calculate the derivative inside the exponent:

\( \frac{d}{dx}(x-x^2) = 1 - 2x \)

So,

\( v'(x) = e^{x(1-x)}(1 - 2x) \)

Substitute \( v(x) \) and \( v'(x) \) back into the expression for \( f'(x) \):

\( f'(x) = (1)(e^{x(1-x)}) + x(e^{x(1-x)}(1-2x)) \)

Simplify:

\( f'(x) = e^{x(1-x)} + x(1-2x)e^{x(1-x)} \) \( f'(x) = e^{x(1-x)}(1 + x(1-2x)) \)

Further simplify:

\( f'(x) = e^{x(1-x)}(1 + x - 2x^2) \) \( f'(x) = e^{x(1-x)}(1 + x - 2x^2) \) \( f'(x) = e^{x(1-x)}(-2x^2 + x + 1) \)

Step 3: Find the critical points by solving \( f'(x) = 0 \).

The exponential function \( e^{x(1-x)} \) is never zero, so we set the quadratic part equal to zero:

\( -2x^2 + x + 1 = 0 \)

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -2 \), \( b = 1 \), \( c = 1 \):

\( x = \frac{-1 \pm \sqrt{1^2 - 4(-2)(1)}}{2(-2)} \) \( x = \frac{-1 \pm \sqrt{1 + 8}}{-4} \) \( x = \frac{-1 \pm \sqrt{9}}{-4} \) \( x = \frac{-1 \pm 3}{-4} \)

Calculate the roots:

\( x_1 = \frac{2}{-4} = -\frac{1}{2} \) \( x_2 = \frac{-4}{-4} = 1 \)

Step 4: Test intervals around the critical points.

Check \( f'(x) \) in the intervals \( (-\infty, -\frac{1}{2}) \), \( (-\frac{1}{2}, 1) \), and \( (1, \infty) \).

• Interval \( (-\infty, -\frac{1}{2}) \): Choose \( x = -1 \), \( f'(x) < 0 \) (decreasing).
• Interval \( (-\frac{1}{2}, 1) \): Choose \( x = 0 \), \( f'(x) > 0 \) (increasing).
• Interval \( (1, \infty) \): Choose \( x = 2 \), \( f'(x) < 0 \) (decreasing).

Therefore, the function \( f(x) \) is increasing in the interval \( (-\frac{1}{2}, 1) \).