Applying the Intermediate Value Theorem to show a root exists

Since $f(x) = x^5 + 3x - 2$ is a polynomial, it is continuous for all real numbers, and in particular continuous on the closed interval $[0,1]$. This satisfies the hypothesis needed to apply the Intermediate Value Theorem on this interval.

Evaluate the endpoints: $f(0) = 0^5 + 3(0) - 2 = -2$, and $f(1) = 1^5 + 3(1) - 2 = 1 + 3 - 2 = 2$. Since $f(0) = -2 < 0$ and $f(1) = 2 > 0$, 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 must exist some $c$ in $(0,1)$ such that $f(c) = 0$. This proves the equation $x^5+3x-2=0$ has at least one root in $(0,1)$.