Question

The function $f$ given by $f(x)=(x^{2}-3)e^{x}$ is decreasing on the interval

• A. $(-3,\infty)$
• B. $(1,\infty)$
• C. $(-\infty,1)$
• D. $(-\infty,-3)$
• E. $(-3,1)$

Hint:

Calculus Tip: Because $e^x$ is never zero or negative, you can safely divide both sides of an inequality by it without flipping the sign. This reduces complex exponential inequalities to simple algebra instantly!

Answer 1

Answer: (E)

Concept:
A function is strictly decreasing on an interval where its first derivative is less than zero ($f^{\prime}(x) < 0$). To find this interval, we must calculate the derivative using the Product Rule, set up the inequality, and solve for $x$.

Step 1: Apply the Product Rule to find the derivative.
Given $f(x) = (x^2 - 3)e^x$, we use $(uv)^{\prime} = u^{\prime}v + uv^{\prime}$: $$f^{\prime}(x) = (2x)e^x + (x^2 - 3)e^x$$

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

Step 3: Set up the inequality for a decreasing function.
For the function to be decreasing, $f^{\prime}(x) < 0$: $$e^x(x^2 + 2x - 3) < 0$$ Since the exponential function $e^x$ is always positive for all real numbers, the sign of the entire expression depends solely on the quadratic part: $$x^2 + 2x - 3 < 0$$

Step 4: Factor the quadratic expression.
Find two numbers that multiply to $-3$ and add to $2$: $$(x + 3)(x - 1) < 0$$ The critical boundary points are $x = -3$ and $x = 1$.

Step 5: Determine the valid interval.
Testing a value between the roots (e.g., $x = 0$) gives $(0 + 3)(0 - 1) = -3 < 0$, which satisfies the inequality. Therefore, the function is decreasing between these two roots. Interval: $(-3, 1)$ Hence the correct answer is (E) $(-3,1)$.