Solution Using the Intermediate Value Theorem (IVT)

The given polynomial function is:
$f(x) = 12x^4 - 5x^2 + 6x - 1$ and the given interval is $[-2, 0]$.

Step 1: Compute $f(-2)$

$f(-2) = 12(-2)^4 - 5(-2)^2 + 6(-2) - 1$ $= 12(16) - 5(4) + (-12) - 1$ $= 192 - 20 - 12 - 1$ $= 159$ Thus,
$f(-2) = 159.$

Step 2: Compute $f(0)$

$f(0) = 12(0)^4 - 5(0)^2 + 6(0) - 1$ $= 0 - 0 + 0 - 1$ $= -1$ Thus,
$f(0) = -1.$

Step 3: Apply the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem 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 $f(c) = 0$.

• Here, $f(-2) = 159$ (positive) and $f(0) = -1$ (negative).
• Since $f(x)$ is a polynomial (which is continuous everywhere), the IVT guarantees that there is some $c$ in $(-2, 0)$ where $f(c) = 0$.

Thus, $f(x)$ has a zero in the given interval, so the correct answer is: $\textbf{Yes}$

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Do you need more details or have any questions?

Here are 5 related questions to explore further:

1. What is the formal statement of the Intermediate Value Theorem?
2. How do you use IVT to estimate the zero of a function?
3. Can IVT be used for functions that are not polynomials?
4. How do you find the exact zero of $f(x)$ in the interval numerically?
5. What happens if both $f(a)$ and $f(b)$ have the same sign?

Tip: IVT guarantees a zero exists but does not give its exact value—you may need numerical methods like the bisection method to find it!