Applying the IVT to prove a root exists

Since $f(x) = x^4 - 5x^2 + 1$ is a polynomial, it is continuous everywhere, and in particular on the closed interval $[0,1]$. This satisfies the continuity hypothesis needed to apply the Intermediate Value Theorem on this interval.

Now evaluate the endpoints: $f(0) = 0^4 - 5(0)^2 + 1 = 1$, and $f(1) = 1^4 - 5(1)^2 + 1 = 1 - 5 + 1 = -3$. Since $f(0) = 1 > 0$ and $f(1) = -3 < 0$, the value $N = 0$ lies strictly between $f(0)$ and $f(1)$.

By the Intermediate Value Theorem, since $f$ is continuous on $[0,1]$ and $0$ is between $f(0)$ and $f(1)$, there exists at least one $c \in (0,1)$ such that $f(c) = 0$. Therefore, $f$ has a root in $(0,1)$.

No, they have opposite signs. The function equals one at zero, which is positive, and equals negative three at one, which is negative. This sign change is exactly what allows the Intermediate Value Theorem to guarantee a root between the two points.