Applying the IVT to show a velocity value is attained

Since $v(t) = t^2 - 4t + 5$ is a polynomial, it is continuous everywhere, and in particular on the closed interval $[0,4]$. Evaluating the endpoints gives $v(0) = 5$ and $v(4) = 16 - 16 + 5 = 5$. Since both endpoint values equal $5$ and $3 < 5$, the value $N = 3$ does not lie strictly between $v(0)$ and $v(4)$. Instead, we apply the IVT on sub-intervals: Consider $[0,2]$ where $v(0)=5$ and $v(2) = 4-8+5=1$. Since $3$ is between $1$ and $5$, the IVT guarantees a $c$ in $(0,2)$ with $v(c)=3$. Similarly, consider $[2,4]$ where $v(2) = 1$ and $v(4) = 5$. Since $3$ is between $1$ and $5$, the IVT also guarantees a $c$ in $(2,4)$ with $v(c)=3$.

To find $c$ explicitly, solve $t^2 - 4t + 5 = 3$, i.e. $t^2 - 4t + 2 = 0$. Using the quadratic formula, $t = \dfrac{4 \pm \sqrt{16-8}}{2} = 2 \pm \sqrt{2}$. We have $c = 2-\sqrt{2} \approx 0.586$, which lies in $(0,2)$ and corresponds to the IVT application on $[0,2]$. We also have $c = 2+\sqrt{2} \approx 3.414$, which lies in $(2,4)$ and corresponds to the IVT application on $[2,4]$. Thus both solutions are valid times where the velocity equals $3$ mph.