se the Intermediate Value Theorem to show that there is a solution to the equation
x^3 cos^4 x = 2024
with 15 ≤ x ≤ 16.


The **Intermediate Value Theorem (IVT)** states that if a function \( f(x) \) is continuous on a closed interval \([a, b]\), and if \( f(a) \) and \( f(b) \) have opposite signs (i.e., one is positive and the other is negative), then there exists at least one \( c \in (a, b) \) such that \( f(c) = 0 \).

### Problem Breakdown:
We are tasked with showing that the equation:

\[
x^3 \cos^4(x) = 2024
\]

has a solution for \( x \in [15, 16] \). To apply the IVT, we first define a function and check whether it meets the requirements of the theorem.

### Step 1: Define the Function
Rearrange the equation to define a continuous function \( f(x) \) by subtracting 2024 from both sides:

\[
f(x) = x^3 \cos^4(x) - 2024
\]

Now we need to find if \( f(x) \) changes signs on the interval \( [15, 16] \).

### Step 2: Evaluate the Function at the Endpoints
First, evaluate \( f(x) \) at the endpoints of the interval.

- For \( x = 15 \):
  \[
  f(15) = 15^3 \cos^4(15) - 2024
  \]
  Using an approximation for \( \cos(15) \) (in radians), we compute:

  \[
  \cos(15) \approx -0.7597
  \]

  \[
  \cos^4(15) = (-0.7597)^4 \approx 0.3326
  \]

  So,

  \[
  f(15) = 15^3 \times 0.3326 - 2024 = 3375 \times 0.3326 - 2024 \approx 1121.475 - 2024 = -902.525
  \]

  Thus, \( f(15) \approx -902.525 \).

- For \( x = 16 \):
  \[
  f(16) = 16^3 \cos^4(16) - 2024
  \]
  Using an approximation for \( \cos(16) \) (in radians), we compute:

  \[
  \cos(16) \approx -0.9577
  \]

  \[
  \cos^4(16) = (-0.9577)^4 \approx 0.8411
  \]

  So,

  \[
  f(16) =

Use the Intermediate Value Theorem to show that there is a solution to the equation x^3 cos^4 x = 2024 with 15 ≤ x ≤ 16.

Learn how to apply the Intermediate Value Theorem to prove the existence of a solution to the equation x^3 cos^4(x) = 2024 in the interval [15, 16]. This example involves polynomial and trigonometric functions, suitable for advanced high school and early university calculus students.

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