The problem concerns using the Intermediate Value Theorem (IVT) to determine if the function $f(x) = x^3 - 3x - 1.5$ has a root in the interval $(0, 1)$. Let's solve it step by step:

Step 1: Intermediate Value Theorem Overview

The IVT states: If a function $f(x)$ is continuous on a closed interval $[a, b]$, and if $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.

Step 2: Verify Continuity

1. Continuity of $f(x)$: The function $f(x) = x^3 - 3x - 1.5$ is a polynomial, which is continuous everywhere, including the interval $[0, 1]$.

   Answer to i): $f$ is continuous on $[0, 1]$.

Step 3: Evaluate $f(0)$ and $f(1)$

We now calculate $f(x)$ at the endpoints of the interval:

1. $f(0) = (0)^3 - 3(0) - 1.5 = -1.5$
2. $f(1) = (1)^3 - 3(1) - 1.5 = 1 - 3 - 1.5 = -3.5$

Step 4: Check Opposite Signs

1. The signs of $f(0) = -1.5$ and $f(1) = -3.5$ are the same (both negative). Therefore, $f(x)$ does not satisfy the condition of the IVT (having opposite signs at the endpoints).

Final Answer

Since $f(x)$ does not meet the necessary conditions for the Intermediate Value Theorem, we cannot use IVT to show that $f(x)$ has a root in $(0, 1)$.

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Related Questions to Explore:

1. What is the Intermediate Value Theorem, and why is continuity crucial for it?
2. How would the conclusion change if $f(0)$ and $f(1)$ had opposite signs?
3. Can we graph $f(x) = x^3 - 3x - 1.5$ to verify its behavior in $(0, 1)$?
4. Are there numerical methods (e.g., Newton's Method) to find roots of $f(x)$?
5. How does polynomial continuity ensure $f(x)$ behaves predictably?

Tip:

When applying the Intermediate Value Theorem, always ensure that the function is continuous and check the signs of $f(a)$ and $f(b)$ to confirm they are opposite.