Step 1: Concept Definition:
A function is considered increasing on an interval if its first derivative is positive, meaning \(f'(x)>0\).
Step 2: Methodology:
The first derivative of \(f(x)\) will be computed using the product rule: \((uv)' = u'v + uv'\). Subsequently, the intervals where \(f'(x)>0\) will be determined.
Step 3: Derivation and Analysis:
The provided function is \(f(x) = x^2e^{-x}\).
Applying the product rule for differentiation, 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, \(f'(x)>0\).
\[ x(2-x)e^{-x}>0 \]
Given that \(e^{-x}\) is always positive for all real values of x, the sign of \(f'(x)\) is determined by the term \(x(2-x)\).
Thus, the inequality to solve is:
\[ x(2-x)>0 \]
The critical points are x = 0 and x = 2. The sign of the expression is analyzed across 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 \(f(x)\) exhibits an increasing behavior on the interval (0, 2).