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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First, notice the function is f(x) = x^2 e^{-x}. To locate inflection points, we need where the second derivative f''(x) changes sign and equals zero.

Compute derivatives step by step: f'(x) = derivative of x^2 e^{-x} = e^{-x}(2x - x^2).
Then f''(x) = derivative of e^{-x}(2x - x^2) = e^{-x}[-(2x - x^2) + (2 - 2x)] = e^{-x}(x^2 - 4x + 2).

Set f''(x) = 0 to find candidate inflection points: x^2 - 4x + 2 = 0.
Solve the quadratic: x = [4 ± sqrt(16 - 8)]/2 = [4 ± sqrt(8)]/2 = 2 ± sqrt(2).

Thus the inflection points occur at x = a ± sqrt(b) with a and b integers and b non-negative. Matching the form 2 ± sqrt(2), we identify a = 2 and b = 2.

Therefore, the values are a = 2 and b = 2, giving inflection points at x = 2 ± sqrt(2).

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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