Question textThe function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2 e^{-x} has two inflection points. These occur when x = a\pm\sqrt{b} for non-negative integer values a= Answer 1 Question 26[input] and b= Answer 2 Question 26[input].

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We start by restating the problem: f(x) = x^2 e^{-x} on the real line, and we are asked to identify the inflection points by expressing them as x = a ± sqrt(b) with a and b being non-negative integers.

First, compute the derivatives needed for concavity changes. The function is f(x) = x^2 e^{-x}. Its first derivative is f'(x) = e^{-x}(2x - x^2) = e^{-x} x(2 - x).

Next, differentiate again to get the second derivative: f''(x) = d/dx [e^{-x} x(2 - x)]. Using the product rule and chain rule, f''(x) = e^{-x}[ x^2 - 4x + 2 ]. (This comes from simplifying -x(2 - x) + (2 - x - x) inside the brackets and then factoring e^{-x}.)

Inflection points occur where f''(x) = 0 and the concavity changes sign. Since e^{-x} > 0 for all x, the zeros come from solving x^2 - 4x + 2 = 0.

Solve the quadratic: x^2 - 4x + 2 = 0 => x = [4 ± sqrt(16 - 8)]/2 = [4 ± sqrt(8)]/2 = [4 ± 2*sqrt(2)]/2 = 2 ± sqrt(2).

Express these in the form a ± sqrt(b) with integers a, b ≥ 0. We have a = 2 and b = 2, giving the two inflection points at x = 2 ± sqrt(2).

To verify the sign change (a necessary condition for an inflection point), note that the quadratic x^2 - 4x + 2 has real distinct roots, so its sign changes across each root, implying f''(x) changes sign at x = 2 ± sqrt(2).

Therefore, the required non-negative integers are a = 2 and b = 2, which matches the provided answer values.

Question textThe function f \colon \mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2 e^{-x} has two inflection points. These occur when x = a\pm\sqrt{b} for non-negative integer values a= Answer 1 Question 26[input] and b= Answer 2 Question 26[input].

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