1. The problem states that $f$ is continuous on the interval $[1,5]$ with $f(1) = 5$ and $f(5) = 2$. 2. The Intermediate Value Theorem (IVT) says that if a function is continuous on a closed interval $[a,b]$, then for any value $N$ between $f(a)$ and $f(b)$, there exists some $c$ in $(a,b)$ such that $f(c) = N$. 3. Here, $f(1) = 5$ and $f(5) = 2$, so the values of $f$ at the endpoints are 5 and 2. 4. The value $1$ is not between $5$ and $2$ because $1 < 2 < 5$ or $1$ is less than both endpoint values. 5. Since $1$ is not between $f(1)$ and $f(5)$, the IVT does not guarantee a $c$ in $(1,5)$ such that $f(c) = 1$. 6. Therefore, the statement "there is a $c$ in $(1,5)$ with $f(c) = 1$" is false. Final answer: B. False