91Ó°ÊÓ

We will use the product rule to find the first derivative. The product rule states that if a function is the product of two functions (u(x) and v(x)), then its derivative is given by: \( \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) \) Given function: \( g(x) = x^2 e^{-x} \) Let \(u(x) = x^2\) and \(v(x) = e^{-x}\) Now, finding the derivatives \(u'(x)\) and \(v'(x)\): \(u'(x) = \frac{d}{dx}(x^2) = 2x\) \(v'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x}\) Now, applying the product rule: \(g'(x) = u'(x)v(x) + u(x)v'(x) = 2xe^{-x} + x^2 (-e^{-x}) = e^{-x}(2x-x^2)\)

Now, we will find the second derivative, g''(x), by computing the derivative of g'(x). To do that, we will again use the product rule: Given \(g'(x) = e^{-x}(2x-x^2)\) Let \(u_1(x) = e^{-x}\) and \(v_1(x) = 2x-x^2\) Now, finding the derivatives \(u_1'(x)\) and \(v_1'(x)\): \(u_1'(x) = \frac{d}{dx}(e^{-x}) = -e^{-x}\) \(v_1'(x) = \frac{d}{dx}(2x-x^2) = 2-2x\) Now, applying the product rule: \(g''(x) = u_1'(x)v_1(x) + u_1(x)v_1'(x) = -e^{-x}(2x-x^2) + e^{-x}(2-2x) = e^{-x}(-x^2+4x-2)\)

Now, we will find the intervals where the second derivative is positive (concave upward) or negative (concave downward). \(g''(x) = e^{-x}(-x^2+4x-2)\) Since \(e^{-x}\) is always positive for any value of x, the sign of g''(x) depends on the quadratic function, \(-x^2 + 4x - 2\). To find its sign, we need to find its roots: \(-x^2 + 4x - 2 = 0 \) We can use the quadratic formula to find the roots: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For this quadratic, a = -1, b = 4, and c = -2. By applying the quadratic formula: \(x = \frac{-4 \pm \sqrt{4^2 - 4(-1)(-2)}}{-2}\) \(x = \frac{-4 \pm \sqrt{16 - 8}}{-2}\) \(x = \frac{-4 \pm \sqrt{8}}{-2}\) \(x = 2 \pm \sqrt{2}\) So, the roots are \(x = 2 - \sqrt{2}\) and \(x = 2 + \sqrt{2}\). Now, we test the intervals between the roots: - For \(x < 2 - \sqrt2\): \(-x^2 + 4x - 2 > 0\), so g''(x) > 0, meaning the function is concave upward in this interval. - For \(2 - \sqrt2 < x < 2 + \sqrt2\): \(-x^2 + 4x - 2 < 0\), so g''(x) < 0, meaning the function is concave downward in this interval. - For \(x > 2 + \sqrt2\): \(-x^2 + 4x - 2 > 0\), so g''(x) > 0, meaning the function is concave upward in this interval.

An inflection point occurs when the concavity of the function changes. From our analysis in step 3, we can see that the concavity changes at \(x = 2 - \sqrt2\) and \(x = 2 + \sqrt2\). Thus, these are our inflection points. To find the corresponding y-values of the inflection points, plug the x-values back into the original function g(x): Inflection point 1: \(g(2-\sqrt2) = (2-\sqrt2)^2 e^{-(2-\sqrt2)}\) Inflection point 2: \(g(2+\sqrt2) = (2+\sqrt2)^2 e^{-(2+\sqrt2)}\) So, the inflection points are \((2-\sqrt2, (2-\sqrt2)^2 e^{-(2-\sqrt2)})\) and \((2+\sqrt2, (2+\sqrt2)^2 e^{-(2+\sqrt2)})\).