Second Derivative Test Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima.

f(x) = x²e⁻ˣ

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First, find the first derivative of the function \( f(x) = x^2 e^{-x} \). Use the product rule, which states that if you have a function \( u(x)v(x) \), its derivative is \( u'(x)v(x) + u(x)v'(x) \). Here, let \( u(x) = x^2 \) and \( v(x) = e^{-x} \).

Calculate the derivatives: \( u'(x) = 2x \) and \( v'(x) = -e^{-x} \). Substitute these into the product rule to find \( f'(x) = 2x e^{-x} + x^2 (-e^{-x}) \). Simplify to get \( f'(x) = e^{-x} (2x - x^2) \).

Set the first derivative \( f'(x) = e^{-x} (2x - x^2) \) equal to zero to find the critical points. Since \( e^{-x} \) is never zero, solve \( 2x - x^2 = 0 \). Factor the equation to get \( x(2 - x) = 0 \), giving critical points at \( x = 0 \) and \( x = 2 \).

Next, find the second derivative \( f''(x) \) to apply the Second Derivative Test. Differentiate \( f'(x) = e^{-x} (2x - x^2) \) using the product rule again. Let \( u(x) = e^{-x} \) and \( v(x) = 2x - x^2 \).

Calculate \( f''(x) = e^{-x}(-1)(2x - x^2) + e^{-x}(2 - 2x) \). Simplify this expression. Evaluate \( f''(x) \) at the critical points \( x = 0 \) and \( x = 2 \) to determine the nature of these points: if \( f''(x) > 0 \), it's a local minimum; if \( f''(x) < 0 \), it's a local maximum.

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Critical Points

Critical points of a function occur where its first derivative is zero or undefined. These points are essential for identifying potential local maxima and minima, as they represent locations where the function's slope changes. To find critical points, one must differentiate the function and solve for the values of x that satisfy the condition f'(x) = 0.