To use the Intermediate Value Theorem (IVT) to show that the polynomial $f(x) = 2x^4 - 9x^2 + 2$ has a real zero between $x = -2$ and $x = 0$, we need to follow these steps:

Step 1: Evaluate $f(x)$ at the endpoints

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

Since $f(x)$ is a polynomial, it is continuous everywhere, including on the interval $[-2, 0]$.

Let's evaluate $f(-2)$ and $f(0)$.

1. Calculate $f(-2)$: $f(-2) = 2(-2)^4 - 9(-2)^2 + 2$ $= 2 \cdot 16 - 9 \cdot 4 + 2$ $= 32 - 36 + 2$ $= -2$

2. Calculate $f(0)$: $f(0) = 2(0)^4 - 9(0)^2 + 2$ $= 0 - 0 + 2$ $= 2$

Step 2: Check the signs of $f(-2)$ and $f(0)$

• $f(-2) = -2$, which is negative.
• $f(0) = 2$, which is positive.

Since $f(-2)$ and $f(0)$ have opposite signs, by the Intermediate Value Theorem, there must be at least one value $c$ in the interval $(-2, 0)$ such that $f(c) = 0$.

Conclusion

The function $f(x) = 2x^4 - 9x^2 + 2$ has at least one real zero in the interval $(-2, 0)$.

Would you like further details, or do you have any questions about this solution?

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Here are some related questions to deepen your understanding:

1. How does the Intermediate Value Theorem apply to functions that are not polynomials?
2. Can the Intermediate Value Theorem be used to find exact values of zeros?
3. How would you use the IVT to show a zero exists in another interval, like $(0, 2)$?
4. What other methods can be used to approximate the zero in $(-2, 0)$?
5. Can the IVT tell us how many zeros exist in an interval?

Tip: The Intermediate Value Theorem only guarantees the existence of a zero in the interval but does not tell us how many zeros or where they are exactly located.