To verify that the root of the equation $x^3 - 3x + 1 = 0$ lies between 0 and 0.8, we need to check the values of the function $f(x) = x^3 - 3x + 1$ at the endpoints of the interval. If the function changes sign between these points, then by the Intermediate Value Theorem, there is at least one root in this interval.

1. Calculate $f(0)$: $f(0) = 0^3 - 3 \cdot 0 + 1 = 1$

2. Calculate $f(0.8)$: $f(0.8) = (0.8)^3 - 3 \cdot 0.8 + 1 = 0.512 - 2.4 + 1 = -0.888$

Now we observe:

• $f(0) = 1$ (positive)
• $f(0.8) = -0.888$ (negative)

Since $f(0)$ is positive and $f(0.8)$ is negative, the function $f(x) = x^3 - 3x + 1$ changes sign between $x = 0$ and $x = 0.8$. By the Intermediate Value Theorem, there is at least one root of the equation $x^3 - 3x + 1 = 0$ in the interval $(0, 0.8)$.

Therefore, we have verified that a root of the equation lies between 0 and 0.8.

Would you like any additional details or have any questions?

Here are some related questions for further exploration:

1. What is the exact value of the root in the interval $(0, 0.8)$?
2. How many roots does the equation $x^3 - 3x + 1 = 0$ have in total?
3. Can you find another interval where another root of $x^3 - 3x + 1 = 0$ lies?
4. How would you use Newton's method to approximate the root in the interval $(0, 0.8)$?
5. What does the graph of $y = x^3 - 3x + 1$ look like?
6. How would you determine the intervals where the function $f(x) = x^3 - 3x + 1$ is increasing or decreasing?
7. What are the critical points of the function $f(x) = x^3 - 3x + 1$?
8. How do you determine the concavity and inflection points of the function $f(x) = x^3 - 3x + 1$?

Tip: When verifying roots of a polynomial equation within a certain interval, always check the signs of the function values at the endpoints and apply the Intermediate Value Theorem for a thorough validation.