已知函数 $$f(x) = e^x + ax^2 - x.$$ (1) 当$a = 1$时，讨论...

Analyzing function f(x)

The query involves analyzing the function $f(x) = e^x + ax^2 - x$.

For part (1), with $a=1$, the function becomes $f(x) = e^x + x^2 - x$. We need to discuss its monotonicity. Specifically, we’ll compute the derivative $f'(x) = e^x + 2x - 1$ and check where it is positive or negative to find the intervals where the function is increasing or decreasing.

Let me check that derivative and proceed with analyzing the sign of $f'(x)$.

Solving inequalities and finding critical points

For the first part, we have the function $g(x) = e^x + 2x - 1$ and want to find its critical points. Solving $g(x) = 0$, we find that $x = 0$ is a root. The derivative $g'(x) = e^x + 2$ is always positive, so $g(x)$ is strictly increasing. Therefore, $x = 0$ is the only root, and $f(x)$ is decreasing for $x < 0$, increasing for $x > 0$, with a minimum at $x = 0$.

For part (2), I'll need to find the range of values of $a$ that satisfy the inequality for all $x \geq 0$. Let's start by analyzing the function $h(x) = e^x - 1 + a x^2 - x - 1/2 x^3$.

Solving for the range of a

Let's analyze the function $h(x)$. Its series expansion for small $x$ gives us an expression $h(x) = x^2(a + 1/2) - (1/3)x^3 + O(x^4)$. For $h(x)$ to be nonnegative near $x = 0$, we need $a + 1/2 \geq 0$, so $a \geq -1/2$. This condition matches the requirement for the second derivative at $x = 0$. However, for small positive $x$, when $a = -1/2$, $h(x)$ becomes negative. So, we have to check further for larger values of $x$.

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Determining borderline condition for a

I’m trying to find the point where minimum occurs for the threshold value $a_0$. First, I set up equations:

1. $h(x_0) = e^{x_0} - 1 + a_0 x_0^2 - x_0 - 1/2 x_0^3 = 0$
2. $h'(x_0) = e^{x_0} + 2a_0 x_0 - 1 - (3/2)x_0^2 = 0$.

After simplifying, I conclude the necessary condition is $a \geq \frac{7 - e^2}{4}$, or approximately -0.0973. This ensures the inequality holds for all $x \geq 0$.

Reasoned for 1m 12s