Question

The function given by $f(x)=x^{3}e^{x}$ is increasing on the interval

• A. $(0,\infty)$
• B. $(3,\infty)$
• C. $(-3,\infty)$
• D. $(-3, 3)$
• E. $(-\infty,-3)$

Hint:

Calculus Tip: While the derivative is technically exactly zero at $x=0$, a function is still considered strictly increasing across that point if the derivative doesn't actually change sign (it stays positive on both sides of $0$).

Answer 1

The Correct Option is C

Concept:
A function $f(x)$ is increasing on an interval if its first derivative $f^{\prime}(x)$ is greater than zero ($f^{\prime}(x) > 0$). To find this interval, we compute the derivative using the Product Rule, set up an inequality, and determine where the expression is positive.

Step 1: Differentiate the function using the Product Rule.
Let $u = x^3$ and $v = e^x$. The Product Rule is $(uv)^{\prime} = u^{\prime}v + uv^{\prime}$. $$f^{\prime}(x) = (3x^2)(e^x) + (x^3)(e^x)$$

Step 2: Factor the derivative.
Factor out the common terms $x^2$ and $e^x$ to simplify the expression: $$f^{\prime}(x) = x^2e^x(3 + x)$$

Step 3: Analyze the signs of the factored components.
For the function to be increasing, we need $x^2e^x(3 + x) > 0$. Let's look at the pieces: 1. $x^2 \ge 0$ for all real numbers (it's strictly positive except at $x=0$). 2. $e^x > 0$ for all real numbers (exponential functions are always positive). Therefore, the sign of the entire derivative relies completely on the binomial $(3 + x)$.

Step 4: Solve the primary inequality.
Since $x^2e^x$ is positive, we simply set the remaining factor greater than zero: $$3 + x > 0$$ $$x > -3$$

Step 5: State the final interval.
The condition $x > -3$ translates to all numbers from $-3$ up to positive infinity. Interval notation: $(-3, \infty)$. Hence the correct answer is (C) $(-3,\infty)$.