Question

The interval, on which the function f(x) = x2e-x is increasing, is equal to

• A. (-$\infty$, $\infty$)
• B. (-$\infty$, 2) $\cup$ (2, $\infty$)
• C. (-2, 0)
• D. (0, 2)

Hint:

To find intervals of increasing/decreasing, always find the first derivative, set it to zero to find critical points, and then test the sign of the derivative in the intervals between these points. Remember that \(e^z\) is always positive.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A function is increasing on an interval where its first derivative is positive, i.e., f'(x)>0.
Step 2: Key Formula or Approach:
We need to find the first derivative of f(x) using the product rule: \((uv)' = u'v + uv'\).
Then, we find the intervals where f'(x)>0.
Step 3: Detailed Explanation:
The given function is \(f(x) = x^2e^{-x}\).
Differentiating f(x) with respect to x using the product rule, where \(u = x^2\) and \(v = e^{-x}\):
\[ f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) \] \[ f'(x) = (2x)e^{-x} + x^2(-e^{-x}) \] \[ f'(x) = e^{-x}(2x - x^2) \] \[ f'(x) = x(2-x)e^{-x} \] For the function to be increasing, we must have f'(x)>0.
\[ x(2-x)e^{-x}>0 \] Since \(e^{-x}\) is always positive for all real x, the sign of f'(x) depends on the term \(x(2-x)\).
So we need to solve the inequality:
\[ x(2-x)>0 \] The critical points are x = 0 and x = 2. We can analyze the sign in the intervals determined by these points:


For x < 0: \(x\) is negative, \((2-x)\) is positive. Product is negative.
For 0 < x < 2: \(x\) is positive, \((2-x)\) is positive. Product is positive.
For x > 2: \(x\) is positive, \((2-x)\) is negative. Product is negative.
The derivative f'(x) is positive when 0 < x < 2.
Step 4: Final Answer:
The function f(x) is increasing on the interval (0, 2).