To show that the equation $x^4 + 7x^3 - 9 = 0$ has at least one positive and one negative solution, we can apply the Intermediate Value Theorem (IVT). The IVT states that if a continuous function $f(x)$ takes on different signs at two points $a$ and $b$ (i.e., $f(a)$ and $f(b)$ have opposite signs), then there exists at least one root $c$ in the interval $(a, b)$ such that $f(c) = 0$.

Let's denote: $f(x) = x^4 + 7x^3 - 9$ This function is continuous for all $x$ because it is a polynomial.

Steps to Use the Intermediate Value Theorem

1. Evaluate $f(x)$ at some positive and negative values to find sign changes:

   • Let's choose $x = 0$: $f(0) = 0^4 + 7 \cdot 0^3 - 9 = -9$ So, $f(0) = -9$, which is negative.

   • Now let's try a positive value, say $x = 2$: $f(2) = 2^4 + 7 \cdot 2^3 - 9 = 16 + 56 - 9 = 63$ Here, $f(2) = 63$, which is positive.

2. Apply the IVT for a positive solution:

   Since $f(0) = -9$ and $f(2) = 63$, there is a sign change between $x = 0$ and $x = 2$. By the Intermediate Value Theorem, there must be at least one root $c$ in the interval $(0, 2)$, meaning there is at least one positive solution.

3. Check a negative value, say $x = -2$: $f(-2) = (-2)^4 + 7 \cdot (-2)^3 - 9 = 16 - 56 - 9 = -49$ Here, $f(-2) = -49$, which is negative.

4. Choose another negative value, say $x = -3$: $f(-3) = (-3)^4 + 7 \cdot (-3)^3 - 9 = 81 - 189 - 9 = -117$ Here, $f(-3) = -117$, which is also negative, indicating no sign change in this interval.

To verify