Applying the IVT to show a root exists

Let $f(x) = 2^x - x^3$. This function is continuous everywhere, since it is a difference of an exponential function and a polynomial, both of which are continuous on all real numbers; in particular, $f$ is continuous on the closed interval $[0,2]$. Evaluate the endpoints: $f(0) = 2^0 - 0^3 = 1 - 0 = 1$, and $f(2) = 2^2 - 2^3 = 4 - 8 = -4$.

Since $f(0) = 1 > 0$ and $f(2) = -4 < 0$, the value $N = 0$ lies strictly between $f(0)$ and $f(2)$. Because $f$ is continuous on $[0,2]$, the Intermediate Value Theorem guarantees there exists at least one $c$ in $(0,2)$ with $f(c) = 0$. This means $2^c - c^3 = 0$, i.e., $2^c = c^3$, so the equation $2^x = x^3$ has a solution in $(0,2)$.

Opposite signs. The argument relied on zero lying strictly between f(0) and f(2), which requires one value to be positive and the other negative. Since f(0) equals one and f(2) equals negative four, they have opposite signs, so zero lies between them. If both endpoint values shared the same sign, zero would not lie between them, and the theorem would not guarantee a root.