To find the maximum value of the function (f(x) = (2x^2 - 1)e^x), follow these steps:

Step 1: Compute the first derivative

Using the product rule (((uv)' = u'v + uv')):

• (u = 2x^2 - 1), (u' = 4x)
• (v = e^x), (v' = e^x)

[f'(x) = 4x e^x + (2x^2 - 1)e^x = e^x(2x^2 + 4x - 1)]

Step 2: Find critical points

Set (f'(x) = 0). Since (e^x \neq 0), solve:
[2x^2 + 4x - 1 = 0]

Quadratic formula: (x = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-2 \pm \sqrt{6}}{2})

Critical points:
(x_1 = \frac{-2 + \sqrt{6}}{2} \approx 0.2245) (local minimum)
(x_2 = \frac{-2 - \sqrt{6}}{2} \approx -2.2245) (local maximum, via second derivative test)

Step 3: Compute the maximum value

Substitute (x_2 = \frac{-2 - \sqrt{6}}{2}) into (f(x)):

[2x^2 - 1 = 2\left(\frac{(-2 - \sqrt{6})^2}{4}\right) - 1 = \frac{4 + 4\sqrt{6} + 6}{2} - 1 = 5 + 2\sqrt{6} - 1 = 4 + 2\sqrt{6}]

[e^x = e^{\frac{-2 - \sqrt{6}}{2}} = e^{-1 - \frac{\sqrt{6}}{2}}]

[f(x_2) = (4 + 2\sqrt{6})e^{-1 - \frac{\sqrt{6}}{2}} = \frac{2(2 + \sqrt{6})}{e^{1 + \frac{\sqrt{6}}{2}}}]

Final Answer:
The maximum value is (\boxed{\frac{2(2 + \sqrt{6})}{e^{1 + \frac{\sqrt{6}}{2}}}}) (or approximately (\boxed{0.96}) if numerical form is preferred).

For simplicity, the exact form is:
(\boxed{2(2+\sqrt{6})e^{-\left(1+\frac{\sqrt{6}}{2}\right)}})

(\boxed{2(2+\sqrt{6})e^{-\left(1+\frac{\sqrt{6}}{2}\right)}})

作者声明:本文包含人工智能生成内容。