1. **State the problem:** We want to show that the equation $x^5 - x^3 + 3x - 5 = 0$ has at least one solution in the interval $(1, 2)$ using the Intermediate Value Theorem (IVT). 2. **Recall the Intermediate Value Theorem:** If a function $f$ is continuous on a closed interval $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c$ in $(a, b)$ such that $f(c) = 0$. 3. **Define the function:** Let $$f(x) = x^5 - x^3 + 3x - 5.$$ This is a polynomial function, which is continuous everywhere, including on $[1, 2]$. 4. **Evaluate $f$ at the endpoints:** - At $x=1$: $$f(1) = 1^5 - 1^3 + 3(1) - 5 = 1 - 1 + 3 - 5 = -2.$$ - At $x=2$: $$f(2) = 2^5 - 2^3 + 3(2) - 5 = 32 - 8 + 6 - 5 = 25.$$ 5. **Check the signs:** - $f(1) = -2$ (negative) - $f(2) = 25$ (positive) Since $f(1)$ and $f(2)$ have opposite signs and $f$ is continuous on $[1, 2]$, by the Intermediate Value Theorem, there exists some $c$ in $(1, 2)$ such that $$f(c) = 0.$$ **Final answer:** There is at least one solution to $x^5 - x^3 + 3x - 5 = 0$ in the interval $(1, 2)$.