To show that there is a root of the equation $x^4 + x - 5 = 0$ in the interval $(1, 2)$ using the Intermediate Value Theorem (IVT), let's go through the steps clearly:

1. Define the function:

Let $f(x) = x^4 + x - 5$, which is a continuous function because it is a polynomial. The IVT applies to continuous functions on closed intervals.

2. Evaluate the function at the endpoints of the interval:

You are given that $f(1) = -3$ and $f(2) = 13$. We can verify this by calculating the values:

$f(1) = 1^4 + 1 - 5 = 1 + 1 - 5 = -3$ $f(2) = 2^4 + 2 - 5 = 16 + 2 - 5 = 13$

3. Apply the Intermediate Value Theorem (IVT):

The IVT states that if a function is continuous on a closed interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs (i.e., one is negative and the other is positive), then there must be some $c \in (a, b)$ such that $f(c) = 0$.

In this case:

• $f(1) = -3$ (which is negative),
• $f(2) = 13$ (which is positive).

Since $-3 < 0 < 13$, by the Intermediate Value Theorem, there must be a number $c$ in the interval $(1, 2)$ such that $f(c) = 0$.

4. Conclusion:

Thus, there is a root of the equation $x^4 + x - 5 = 0$ in the interval $(1, 2)$, as guaranteed by the Intermediate Value Theorem.

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Do you want more details on the IVT or anything specific? Here are 5 related questions for deeper understanding:

1. How does the Intermediate Value Theorem apply to other types of functions besides polynomials?
2. Can you explain why continuity is important for the IVT?
3. How would you numerically approximate the value of the root in $(1, 2)$?
4. Could a function have more than one root in the interval $(1, 2)$?
5. How would you check for roots in other intervals, such as $(0, 1)$ or $(2, 3)$?

Tip: When applying the Intermediate Value Theorem, always ensure that the function is continuous and that the values at the endpoints have opposite signs.