Finding the Increasing Interval for $f(x) = x^2e^{-x}$

To determine the interval on which the function $f(x) = x^2e^{-x}$ is increasing, we need to find the first derivative of the function, $f'(x)$, and identify where $f'(x) > 0$.

Step 1: Calculate the First Derivative

We use the product rule for differentiation, which states that if $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = x^2$ and $v(x) = e^{-x}$.

Then, the derivatives are:

• $u'(x) = \frac{d}{dx}(x^2) = 2x$
• $v'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x}$

Applying the product rule:

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

$f'(x) = 2xe^{-x} - x^2e^{-x}$

Factor out the common terms $xe^{-x}$:

$f'(x) = xe^{-x}(2 - x)$

Step 2: Determine Where the Derivative is Positive

The function $f(x)$ is increasing when $f'(x) > 0$.

$xe^{-x}(2 - x) > 0$

Since $e^{-x}$ is always positive for any real value of $x$, the sign of $f'(x)$ depends on the signs of the factors $x$ and $(2 - x)$.

We find the critical points by setting $f'(x) = 0$: $xe^{-x}(2 - x) = 0$ This gives us $x = 0$ and $2 - x = 0 \implies x = 2$. These critical points divide the number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

Step 3: Analyze the Sign of $f'(x)$ in Each Interval

We test a value from each interval to check the sign of $f'(x)$:

• Interval $(-\infty, 0)$: Let's pick $x = -1$. $f'(-1) = (-1)e^{-(-1)}(2 - (-1)) = (-1)e^1(3) = -3e$. Since $-3e < 0$, the function is decreasing in this interval.
• Interval $(0, 2)$: Let's pick $x = 1$. $f'(1) = (1)e^{-1}(2 - 1) = (1)e^{-1}(1) = e^{-1}$. Since $e^{-1} > 0$, the function is increasing in this interval.
• Interval $(2, \infty)$: Let's pick $x = 3$. $f'(3) = (3)e^{-3}(2 - 3) = (3)e^{-3}(-1) = -3e^{-3}$. Since $-3e^{-3} < 0$, the function is decreasing in this interval.

Conclusion

The function $f(x) = x^2e^{-x}$ is increasing on the interval where $f'(x) > 0$. Based on our analysis, this occurs in the interval $(0, 2)$.