Applying the IVT to prove a root exists

Since $f(x) = e^x - 3x$ is a sum/difference of the exponential function $e^x$ and the polynomial $-3x$, both of which are continuous for all real numbers, $f$ is continuous everywhere, and in particular on the closed interval $[0,2]$. This satisfies the hypothesis needed to apply the IVT.

Evaluate the endpoints: $f(0) = e^0 - 3(0) = 1 - 0 = 1 > 0$, and $f(2) = e^2 - 6 \approx 7.389 - 6 = 1.389 > 0$. Since both endpoint values are positive, we cannot directly apply the IVT to the interval $[0,2]$. However, by checking an interior point, $f(1) = e^1 - 3 \approx 2.718 - 3 = -0.282 < 0$, we discover a sign change. Since $f$ is continuous on $[0,1]$ with $f(0) > 0$ and $f(1) < 0$, the IVT guarantees some $c \in (0,1) \subset (0,2)$ with $f(c) = 0$, so a root of $e^x - 3x = 0$ exists in $(0,2)$.

The sign change was located between zero and one, since f of zero was positive and f of one was negative. Although the endpoints zero and two both gave positive values, checking the midpoint one revealed the needed sign change. This smaller interval, zero to one, is where the IVT guarantees the root lies.