Applying the IVT to prove a root exists

Since $f(x) = x^2 + \frac{1}{x+3}$ is a sum of a polynomial and a rational function, it is continuous everywhere except where $x + 3 = 0$, i.e., at $x = -3$. Since $-3$ is not in $[-1, 2]$, $f$ is continuous on the entire closed interval $[-1, 2]$.

Now evaluate the endpoints: $f(-1) = (-1)^2 + \frac{1}{-1+3} = 1 + \frac{1}{2} = 1.5$, and $f(2) = 2^2 + \frac{1}{2+3} = 4 + \frac{1}{5} = 4.2$.

Since $f$ is continuous on $[-1,2]$ and we need $f(c) = 5$, the Intermediate Value Theorem applies only if $5$ lies between $f(-1)$ and $f(2)$. However, $f(-1) = 1.5$ and $f(2) = 4.2$, so $5$ is not between these endpoint values. Therefore, the IVT cannot guarantee the existence of a $c$ in $(-1, 2)$ such that $f(c) = 5$.

The Intermediate Value Theorem fails because both endpoint values are smaller than five: $f(-1) = 1.5$ and $f(2) = 4.2$. Since the target value of $5$ is larger than both endpoint values, it is not bracketed between them. The IVT requires the target value to lie strictly between the endpoint values to guarantee a solution exists on the interval.