Question 64

Define the continuous and infinitely differentiable function:

Compute its derivatives:

Since $e^x > 0$ for all real $x$, we have $f''(x) = e^x + 1 > 1 > 0$ everywhere.
Because $f''(x) > 0$ for all $x \in \mathbb{R}$, $f(x)$ is strictly convex, and its first derivative $f'(x)$ is strictly increasing.

Find the critical point where $f'(x) = 0$:

• $f'(0) = e^0 + 0 = 1 > 0$
• $f'(-1) = e^{-1} - 1 = \frac{1}{e} - 1 \approx 0.368 - 1 = -0.632 < 0$
  By the Intermediate Value Theorem, $f'(x) = 0$ has a unique root $x_0 \in (-1, 0)$.
  Since $f$ is strictly convex, $x_0$ is the unique global minimum of $f(x)$ on $\mathbb{R}$.

Evaluate the function at the minimum $x_0$ and the interval boundaries:

1. At the minimum $x_0 \in (-1, 0)$:
   Since $x_0 < 0$, $e^{x_0} < 1$, and since $|x_0| < 1$, $0.5x_0^2 < 0.5$.
   Therefore, $f(x_0) = e^{x_0} + 0.5x_0^2 - 2 < 1 + 0.5 - 2 = -0.5 < 0$.
   The minimum value is strictly negative.
2. At $x = -5$:
   $$f(-5) = e^{-5} + 0.5(-5)^2 - 2 = e^{-5} + 12.5 - 2 = 10.5 + e^{-5} > 0$$
3. At $x = 5$:
   $$f(5) = e^5 + 0.5(5)^2 - 2 = e^5 + 10.5 > 0$$

Because $f(-5) > 0$, $f(x_0) < 0$, and $f$ is strictly decreasing on $[-5, x_0]$, there is exactly one root in $[-5, x_0]$.
Because $f(x_0) < 0$, $f(5) > 0$, and $f$ is strictly increasing on $[x_0, 5]$, there is exactly one root in $[x_0, 5]$.
Thus, $f(x) = 0$ has exactly 2 roots in the range $[-5, 5]$. This matches option C.