The equation given is:

$\cos(\pi x) - 7x^3 + 5 = 0$

We are asked to use the Intermediate Value Theorem (IVT) to show that there is a solution to this equation in the interval $(0,1)$.

Steps for applying the Intermediate Value Theorem:

1. Verify the conditions of IVT:

   • The function $f(x) = \cos(\pi x) - 7x^3 + 5$ must be continuous on the interval $[0, 1]$.
     • The cosine function is continuous for all real numbers.
     • The polynomial $-7x^3 + 5$ is continuous for all real numbers.
     • Since both components are continuous, $f(x)$ is continuous on $[0, 1]$.

2. Check the values of $f(x)$ at the endpoints of the interval:

   • Compute $f(0)$ and $f(1)$.

   $f(0) = \cos(0) - 7(0)^3 + 5 = 1 - 0 + 5 = 6$

   $f(1) = \cos(\pi) - 7(1)^3 + 5 = -1 - 7 + 5 = -3$

3. Apply the Intermediate Value Theorem:

   • The IVT states that if $f(x)$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there is at least one $c \in (a, b)$ such that $f(c) = 0$.
   • In this case, $f(0) = 6$ and $f(1) = -3$, which have opposite signs.

Since $f(x)$ is continuous on $[0, 1]$ and the values at the endpoints have opposite signs, by the Intermediate Value Theorem, there is at least one solution to the equation $f(x) = 0$ in the interval $(0, 1)$.

Would you like further details or have any other questions?

Here are five related questions:

1. How do you use the Intermediate Value Theorem for other types of equations?
2. What are other methods to find solutions to equations besides the IVT?
3. How can you approximate the solution to the equation numerically?
4. What role does the continuity of a function play in solving equations?
5. How does the concept of limits relate to the Intermediate Value Theorem?

Tip: When applying the IVT, always check that the function is continuous and that the endpoints have different signs to guarantee a solution exists within the interval.