Question

Find the sub-interval of \[ (0,\infty) \] in which \[ f(x)=x^2e^{-x} \] is increasing.

Hint:

Exponential terms like \( e^{-x} \) or \( e^{x} \) can never be negative or zero for any real \( x \). You can divide them out of inequalities safely without changing the inequality sign direction.

Answer 1

Concept: For a continuous function \( f(x) \) to be increasing on an interval, its first derivative must be strictly greater than zero (\( f'(x) > 0 \)) within that interval. We will find \( f'(x) \) using the product rule of differentiation: \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \]

Step 1: Compute the first derivative \( f'(x) \).
The function is given by: \[ f(x) = x^2 e^{-x} \] Let \( u(x) = x^2 \) and \( v(x) = e^{-x} \). Differentiating both with respect to \( x \): \[ u'(x) = 2x \quad \text{and} \quad v'(x) = -e^{-x} \] Applying the product rule: \[ f'(x) = (2x)(e^{-x}) + (x^2)(-e^{-x}) \] Factoring out the common exponential term \( e^{-x} \) and \( x \): \[ f'(x) = e^{-x}(2x - x^2) = x e^{-x}(2 - x) \]

Step 2: Set up the inequality for an increasing function.
For \( f(x) \) to be strictly increasing, we set: \[ f'(x) > 0 \quad \Rightarrow \quad x e^{-x}(2 - x) > 0 \]

Step 3: Analyze the signs of the individual factors over the domain \( (0, \infty) \).
We are given that \( x \in (0, \infty) \), which means:
• \( x > 0 \) is always positive.
• \( e^{-x} > 0 \) is strictly positive for all real values of \( x \). Since \( x \) and \( e^{-x} \) are both positive, the sign of the overall derivative depends solely on the remaining factor \( (2 - x) \): \[ 2 - x > 0 \quad \Rightarrow \quad 2 > x \quad \Rightarrow \quad x < 2 \]

Step 4: Intersect the inequality with the given domain.
Combining our result \( x < 2 \) with the specified domain constraint \( x > 0 \), we obtain the open interval: \[ 0 < x < 2 \quad \Rightarrow \quad x \in (0, 2) \] Thus, the function is increasing on the sub-interval \( (0, 2) \).