Question

The interval in which $y = x^2e^{2x}$ is increasing is

• A. $(-\infty,-1)$
• B. $(-1,\infty)$
• C. $(-\infty,-1)\cup(0,\infty)$
• D. $(-\infty,0)\cup(1,\infty)$

Answer 1

Answer: (C)

$(-\infty,-1)\cup(0,\infty)$

Answer 2

To determine the interval where the function $y = x^2e^{2x}$ is increasing, we need to find its first derivative, $\frac{dy}{dx}$, and analyze its sign.

1. Find the first derivative: We use the product rule, $(uv)' = u'v + uv'$, where $u = x^2$ and $v = e^{2x}$. $u' = \frac{d}{dx}(x^2) = 2x$ $v' = \frac{d}{dx}(e^{2x}) = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$

   So, $\frac{dy}{dx} = (2x)(e^{2x}) + (x^2)(2e^{2x})$ $\frac{dy}{dx} = 2xe^{2x} + 2x^2e^{2x}$ Factor out the common terms, $2xe^{2x}$: $\frac{dy}{dx} = 2xe^{2x}(1 + x)$

2. Find critical points: Set $\frac{dy}{dx} = 0$ to find the critical points: $2xe^{2x}(1 + x) = 0$ Since $e^{2x}$ is always positive for all real $x$, we only need to consider the other factors: $2x(1 + x) = 0$ This equation yields two critical points: $2x = 0 \Rightarrow x = 0$ $1 + x = 0 \Rightarrow x = -1$

3. Analyze the sign of $\frac{dy}{dx}$: The critical points $x = -1$ and $x = 0$ divide the real number line into three intervals: $(-\infty, -1)$, $(-1, 0)$, and $(0, \infty)$. We test the sign of $\frac{dy}{dx}$ in each interval. Remember that the sign of $\frac{dy}{dx}$ is determined by the sign of $2x(1+x)$ since $e^{2x} > 0$.

   • Interval $(-\infty, -1)$: Choose a test value, e.g., $x = -2$. $2(-2)(1 + (-2)) = -4(-1) = 4$. Since $4 > 0$, $\frac{dy}{dx} > 0$ in this interval. Thus, $y$ is increasing.

   • Interval $(-1, 0)$: Choose a test value, e.g., $x = -0.5$. $2(-0.5)(1 + (-0.5)) = -1(0.5) = -0.5$. Since $-0.5 < 0$, $\frac{dy}{dx} < 0$ in this interval. Thus, $y$ is decreasing.

   • Interval $(0, \infty)$: Choose a test value, e.g., $x = 1$. $2(1)(1 + 1) = 2(2) = 4$. Since $4 > 0$, $\frac{dy}{dx} > 0$ in this interval. Thus, $y$ is increasing.

4. Conclusion: The function $y = x^2e^{2x}$ is increasing when $\frac{dy}{dx} > 0$. This occurs in the intervals $(-\infty, -1)$ and $(0, \infty)$. The union of these intervals is $(-\infty, -1) \cup (0, \infty)$.