1. **Problem:** Use the Intermediate Value Theorem (IVT) to show that the function $$f(x) = e^{x/4} + \frac{1}{8}x^2 - 4$$ has a root in the interval $[0,4]$. 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. **Check continuity:** The function $f(x)$ is composed of exponential and polynomial terms, both continuous everywhere, so $f$ is continuous on $[0,4]$. 4. **Evaluate $f(0)$:** $$f(0) = e^{0/4} + \frac{1}{8} \cdot 0^2 - 4 = 1 + 0 - 4 = -3$$ 5. **Evaluate $f(4)$:** $$f(4) = e^{4/4} + \frac{1}{8} \cdot 4^2 - 4 = e^{1} + \frac{1}{8} \cdot 16 - 4 = e + 2 - 4 = e - 2$$ Using $e \approx 2.718$, $$f(4) \approx 2.718 - 2 = 0.718 > 0$$ 6. **Sign check:** At $x=0$, $f(0) = -3 < 0$ and at $x=4$, $f(4) \approx 0.718 > 0$. 7. **Conclusion:** Since $f$ is continuous on $[0,4]$ and $f(0)$ and $f(4)$ have opposite signs, by the Intermediate Value Theorem, there exists some $c \in (0,4)$ such that $$f(c) = 0$$ which means $f$ has at least one root in $[0,4]$. **Final answer:** There is at least one root of $f(x)$ in the interval $[0,4]$ by the Intermediate Value Theorem.