1. **State the problem:** Use the Intermediate Value Theorem (IVT) to show there is a root of the equation $$e^x = 9 - 8x$$ in the interval $$(0,1)$$. 2. **Rewrite the equation:** Define a function $$f(x) = e^x - 9 + 8x$$. Finding a root of the original equation is equivalent to finding $$x$$ such that $$f(x) = 0$$. 3. **Check continuity:** The function $$f(x) = e^x - 9 + 8x$$ is continuous for all real numbers because exponential and polynomial functions are continuous everywhere. 4. **Evaluate at the endpoints:** - At $$x=0$$: $$f(0) = e^0 - 9 + 8(0) = 1 - 9 + 0 = -8$$. - At $$x=1$$: $$f(1) = e^1 - 9 + 8(1) = e - 9 + 8 = e - 1$$. Since $$e \approx 2.718$$, $$f(1) \approx 2.718 - 1 = 1.718 > 0$$. 5. **Apply the Intermediate Value Theorem:** Since $$f(0) = -8 < 0$$ and $$f(1) \approx 1.718 > 0$$, and $$f$$ is continuous on $[0,1]$, by the IVT there exists some $$c \in (0,1)$$ such that $$f(c) = 0$$. 6. **Conclusion:** There is at least one root of the equation $$e^x = 9 - 8x$$ in the interval $$(0,1)$$.