This problem asks to use the Intermediate Value Theorem (IVT) to show that there is a root of the equation $e^x = 3 - 2x$ in the interval $(0, 1)$.

Step-by-Step Breakdown:

1. Rearrange the equation into a function form: $f(x) = e^x - 3 + 2x.$ The task now is to check for values of $f(x)$ at the endpoints of the interval $[0, 1]$.

2. Evaluate $f(x)$ at $x = 0$: $f(0) = e^0 - 3 + 2(0) = 1 - 3 = -2.$ So, $f(0) = -2$.

3. Evaluate $f(x)$ at $x = 1$: $f(1) = e^1 - 3 + 2(1) = e - 3 + 2 = e - 1.$ Since $e \approx 2.718$, we have: $f(1) = 2.718 - 1 = 1.718.$ So, $f(1) = 1.718 \approx 1.72$.

4. Apply the Intermediate Value Theorem (IVT): The IVT states that if a function is continuous on a closed interval and the function takes values of opposite signs at the endpoints of the interval, then there must be at least one root (where $f(x) = 0$) in the interval.

   • $f(x)$ is continuous because $e^x$, $3$, and $2x$ are continuous functions.
   • $f(0) = -2$ and $f(1) = 1.72$. Since one value is negative and the other is positive, by the IVT, there must be some $c \in (0, 1)$ such that $f(c) = 0$.

Corrections for the boxes:

• $f(0) = -2$ (the value you entered, 3, is incorrect).
• $f(1) = 1.72$ (the correct value, not 1).

Thus, there is a root of the equation in the interval $(0, 1)$ by the Intermediate Value Theorem.

Would you like further details on any part of this explanation?

Here are some related questions:

1. How does the Intermediate Value Theorem work?
2. Can the IVT guarantee more than one root in an interval?
3. Why is the continuity of a function important for applying IVT?
4. What is the significance of the function's values at the endpoints of the interval?
5. How could the function be solved analytically or numerically for the root?

Tip: The Intermediate Value Theorem only guarantees the existence of a root in the interval but doesn't provide its exact location—numerical methods may be needed to approximate the root.