Step 1: Concept Definition:
A function is increasing on an interval if its first derivative is positive, indicated by f'(x)>0.
Step 2: Methodology:
Calculate the first derivative of f(x) using the product rule: (uv)' = u'v + uv'.
Determine the intervals where f'(x)>0.
Step 3: Detailed Calculation:
The function provided is \(f(x) = x^2e^{-x}\).
Differentiate f(x) with respect to x using the product rule, with \(u = x^2\) 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, f'(x) must be greater than 0.
\[ x(2-x)e^{-x}>0 \]
Since \(e^{-x}\) is always positive for all real x, the sign of f'(x) is determined by the term \(x(2-x)\).
Solve the inequality:
\[ x(2-x)>0 \]
The critical points are x = 0 and x = 2. Analyze the sign in the intervals defined by these points:
For x<0: \(x\) is negative, \((2-x)\) is positive. The product is negative.
For 0<x<2: \(x\) is positive, \((2-x)\) is positive. The product is positive.
For x>2: \(x\) is positive, \((2-x)\) is negative. The product is negative.
The derivative f'(x) is positive in the interval 0<x<2.
Step 4: Conclusion:
The function f(x) is increasing on the interval (0, 2).