Analyzing the provided exercises:

Exercise 43:

The function is $f(x) = x^3 + 2x^2 - 1$. The task is to explain why there exists at least one real number $c$ between $-2$ and $1$ such that $f(c) = -0.5$.

Solution:

This uses the Intermediate Value Theorem (IVT), which states:
If a function $f$ is continuous on a closed interval $[a, b]$, and if $f(a) \leq k \leq f(b)$ (or $f(b) \leq k \leq f(a)$), then there exists at least one $c \in [a, b]$ such that $f(c) = k$.

1. Verify continuity: The function $f(x) = x^3 + 2x^2 - 1$ is a polynomial, which is continuous everywhere on $\mathbb{R}$, including the interval $[-2, 1]$.
2. Compute $f(-2)$ and $f(1)$: $f(-2) = (-2)^3 + 2(-2)^2 - 1 = -8 + 8 - 1 = -1$ $f(1) = (1)^3 + 2(1)^2 - 1 = 1 + 2 - 1 = 2$ Thus, $f(-2) = -1$ and $f(1) = 2$.
3. Check if $-0.5$ lies between $f(-2)$ and $f(1)$:
   Since $-1 \leq -0.5 \leq 2$, by the Intermediate Value Theorem, there exists at least one ( c \in