The question is asking to apply the Intermediate Value Theorem (IVT) to the given function $f(x) = 2x^3 + 5x^2 + 8$ on the interval $[0,1]$. The IVT states that if a function is continuous on a closed interval, and if $f(a) \neq f(b)$, then for any $k$ between $f(a)$ and $f(b)$, there exists a number $c$ in the interval $[a, b]$ such that $f(c) = k$.

Let's apply the steps:

1. First, calculate $f(0)$: $f(0) = 2(0)^3 + 5(0)^2 + 8 = 8$

2. Next, calculate $f(1)$: $f(1) = 2(1)^3 + 5(1)^2 + 8 = 2 + 5 + 8 = 15$

Now, the values of $f(x)$ at the endpoints of the interval $[0, 1]$ are:

• $f(0) = 8$
• $f(1) = 15$

By the IVT, any value $k$ that lies between $f(0) = 8$ and $f(1) = 15$ will have a corresponding $c$ in the interval $[0, 1]$ such that $f(c) = k$.

Thus, the values of $k$ that satisfy the Intermediate Value Theorem are: $8 \leq k \leq 15$

This is the range for $k$.

Would you like further details on how the Intermediate Value Theorem works or on any part of this solution?

Here are five related questions to deepen your understanding:

1. What happens if $k$ is outside the interval $[8, 15]$?
2. Can we apply the IVT to a discontinuous function?
3. What does the IVT guarantee about the number of solutions for $f(c) = k$?
4. How can the IVT be used to approximate solutions to equations?
5. How can you verify that a function is continuous on a given interval?

Tip: The Intermediate Value Theorem only guarantees the existence of a solution, not how many or exactly where it is located.