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) =