Key Concepts and Formulas

• Monotonicity of a Function: A function $f(x)$ is increasing on an interval if $f'(x) > 0$ for all $x$ in that interval, and decreasing if $f'(x) < 0$ for all $x$ in that interval.
• Product Rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.
• Chain Rule: If $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

Step-by-Step Solution

Step 1: Identify the function and prepare for differentiation. We are given the function $f(x) = x e^{x(1-x)}$. Our goal is to find the intervals where this function is increasing or decreasing. To do this, we need to find its derivative, $f'(x)$, and analyze its sign.

Step 2: Differentiate $f(x)$ using the Product Rule. Since $f(x)$ is a product of two functions, $u(x) = x$ and $v(x) = e^{x(1-x)}$, we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$. First, we find the derivatives of $u(x)$ and $v(x)$: $u(x) = x \implies u'(x) = 1$ $v(x) = e^{x(1-x)}$

Step 3: Differentiate $v(x)$ using the Chain Rule. To differentiate $v(x) = e^{x(1-x)}$, we use the chain rule. Let $g(x) = x(1-x) = x - x^2$. Then $v(x) = e^{g(x)}$. $g'(x) = \frac{d}{dx}(x - x^2) = 1 - 2x$. Therefore, $v'(x) = e^{g(x)} \cdot g'(x) = e^{x(1-x)} \cdot (1 - 2x)$.

Step 4: Substitute the derivatives back into the Product Rule. Now we substitute $u'(x)$, $v(x)$, and $v'(x)$ into the product rule formula: $f'(x) = (1) \cdot e^{x(1-x)} + (x) \cdot e^{x(1-x)} (1 - 2x)$ $f'(x) = e^{x(1-x)} + x(1 - 2x) e^{x(1-x)}$

Step 5: Simplify the expression for $f'(x)$. Factor out the common term $e^{x(1-x)}$: $f'(x) = e^{x(1-x)} [1 + x(1 - 2x)]$ $f'(x) = e^{x(1-x)} [1 + x - 2x^2]$ Rearrange the terms inside the bracket into standard quadratic form: $f'(x) = e^{x(1-x)} [-2x^2 + x + 1]$

Step 6: Factor the quadratic expression. To determine the sign of $f'(x)$, we need to factor the quadratic term $-2x^2 + x + 1$. First, factor out $-1$: $-2x^2 + x + 1 = -(2x^2 - x - 1)$ Now, factor the quadratic $2x^2 - x - 1$. We look for two numbers that multiply to $2 \times (-1) = -2$ and add up to $-1$. These numbers are $-2$ and $1$. So, we can rewrite the middle term: $2x^2 - 2x + x - 1$ Factor by grouping: $2x(x - 1) + 1(x - 1)$ $(2x + 1)(x - 1)$ Therefore, the quadratic expression is: $-2x^2 + x + 1 = -(2x + 1)(x - 1)$ Substitute this back into $f'(x)$: $f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$

Step 7: Find the critical points. The critical points are the values of $x$ where $f'(x) = 0$. Since $e^{x(1-x)}$ is always positive for all $x \in \mathbb{R}$, we only need to set the other factor to zero: $-(2x + 1)(x - 1) = 0$ This implies either $2x + 1 = 0$ or $x - 1 = 0$.

• $2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$
• $x - 1 = 0 \implies x = 1$ So, the critical points are $x = -\frac{1}{2}$ and $x = 1$.

Step 8: Analyze the sign of $f'(x)$ on a number line. We place the critical points $-\frac{1}{2}$ and $1$ on a number line. These points divide the number line into three intervals: $\left(-\infty, -\frac{1}{2}\right)$, $\left(-\frac{1}{2}, 1\right)$, and $\left(1, \infty\right)$. We need to test a value of $x$ from each interval to determine the sign of $f'(x) = -e^{x(1-x)} (2x + 1)(x - 1)$. Remember that $e^{x(1-x)}$ is always positive, so we only need to analyze the sign of $-(2x + 1)(x - 1)$.

• Interval 1: $x < -\frac{1}{2}$ (e.g., choose $x = -1$)

  • $2x + 1 = 2(-1) + 1 = -1$ (negative)
  • $x - 1 = -1 - 1 = -2$ (negative)
  • $(2x + 1)(x - 1) = (-)(-)$ = $(+)$
  • $f'(x) = - (\text{positive term}) (\text{positive}) = (-)$
  • Since $f'(x) < 0$, $f(x)$ is decreasing in $\left(-\infty, -\frac{1}{2}\right)$.

• Interval 2: $-\frac{1}{2} < x < 1$ (e.g., choose $x = 0$)

  • $2x + 1 = 2(0) + 1 = 1$ (positive)
  • $x - 1 = 0 - 1 = -1$ (negative)
  • $(2x + 1)(x - 1) = (+)(-)$ = $(-)$
  • $f'(x) = - (\text{positive term}) (\text{negative}) = (+)$
  • Since $f'(x) > 0$, $f(x)$ is increasing in $\left(-\frac{1}{2}, 1\right)$.

• Interval 3: $x > 1$ (e.g., choose $x = 2$)

  • $2x + 1 = 2(2) + 1 = 5$ (positive)
  • $x - 1 = 2 - 1 = 1$ (positive)
  • $(2x + 1)(x - 1) = (+)(+)$ = $(+)$
  • $f'(x) = - (\text{positive term}) (\text{positive}) = (-)$
  • Since $f'(x) < 0$, $f(x)$ is decreasing in $\left(1, \infty\right)$.

Step 9: Determine which option is correct. Based on our analysis, $f(x)$ is increasing in $\left(-\frac{1}{2}, 1\right)$.

Common Mistakes & Tips

• Forgetting the Chain Rule: When differentiating $e^{x(1-x)}$, remember to multiply by the derivative of the exponent.
• Sign Errors: Be careful with the negative sign when factoring and analyzing the sign of $f'(x)$.
• Incorrect Factoring: Double-check your factoring of the quadratic expression.

Summary

We found the derivative of $f(x) = xe^{x(1-x)}$ using the product and chain rules. We then factored the derivative to find the critical points and analyzed the sign of the derivative in the intervals defined by these critical points. This analysis showed that $f(x)$ is increasing in the interval $\left(-\frac{1}{2}, 1\right)$.

The final answer is \boxed{increasing in $\left(-\frac{1}{2}, 1\right)$} which corresponds to option (A).