Understanding Function Increase and Decrease

A function $y = f(x)$ is said to be increasing on an interval if for any two points $x_1$ and $x_2$ in that interval, $x_1 < x_2$ implies $f(x_1) \le f(x_2)$. For differentiable functions, this condition is met when the first derivative $f'(x)$ is positive in the interval.

In this problem, we are given the function $y = x^2e^{-x}$. To find the interval where this function is increasing, we need to calculate its first derivative, $y'$, and find the intervals where $y' > 0$.

Calculating the First Derivative

We use the product rule for differentiation, which states that if $y = u(x)v(x)$, then $y' = u'(x)v(x) + u(x)v'(x)$. Here, let $u(x) = x^2$ and $v(x) = e^{-x}$.

• The derivative of $u(x) = x^2$ is $u'(x) = \frac{d}{dx}(x^2) = 2x$.
• The derivative of $v(x) = e^{-x}$ requires the chain rule. Let $w = -x$, then $v = e^w$. $\frac{dv}{dx} = \frac{dv}{dw} \cdot \frac{dw}{dx} = e^w \cdot (-1) = -e^{-x}$. So, $v'(x) = -e^{-x}$.

Applying the product rule:

$\qquad y' = u'(x)v(x) + u(x)v'(x)$

$\qquad y' = (2x)(e^{-x}) + (x^2)(-e^{-x})$

$\qquad y' = 2xe^{-x} - x^2e^{-x}$

We can factor out $e^{-x}$ and $x$ from the expression:

$\qquad y' = e^{-x}(2x - x^2)$

$\qquad y' = xe^{-x}(2 - x)$

Finding Critical Points

Critical points are the points where the derivative $y'$ is zero or undefined. The derivative $y' = xe^{-x}(2 - x)$ is defined for all real numbers $x$. So, we only need to find where $y' = 0$.

Set the derivative equal to zero:

$\qquad xe^{-x}(2 - x) = 0$

Since $e^{-x}$ is always positive for all real values of $x$, the equation $xe^{-x}(2 - x) = 0$ is satisfied if either $x = 0$ or $(2 - x) = 0$.

• If $x = 0$, then $x=0$.
• If $2 - x = 0$, then $x = 2$.

The critical points are $x = 0$ and $x = 2$. These points divide the number line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

Analyzing the Sign of the Derivative

To determine where the function is increasing, we test the sign of $y' = xe^{-x}(2 - x)$ in each of the intervals defined by the critical points.

From the sign analysis, we see that $y' > 0$ in the interval $(0, 2)$. This means the function $y = x^2e^{-x}$ is increasing on the interval $(0, 2)$.

Conclusion

The function $y = x^2e^{-x}$ is increasing on the interval $(0, 2)$.

Revision Table: Function Increasing/Decreasing

Additional Information: Exponential Function Properties

The exponential function $e^u$ is always positive for any real number $u$. This is why $e^{-x} > 0$ for all $x$. Understanding this property is crucial when analyzing the sign of derivatives involving exponential terms.

When analyzing the sign of a derivative like $y' = xe^{-x}(2 - x)$, we look at the sign of each factor. Since $e^{-x}$ is always positive, its sign does not affect the overall sign of $y'$. The sign of $y'$ is determined solely by the signs of the other factors, $x$ and $(2 - x)$.