Ivt Root Check 55F991

1. **State the problem:** We want to determine if the equation $$x^3 - 3x - 0.9 = 0$$ has at least one root in the interval $$(0,1)$$ 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 $$f(x) = x^3 - 3x - 0.9$$. 4. **Check continuity:** Since $$f(x)$$ is a polynomial, it is continuous everywhere, including on $$[0,1]$$. 5. **Evaluate $$f$$ at the endpoints:** $$f(0) = 0^3 - 3(0) - 0.9 = -0.9$$ $$f(1) = 1^3 - 3(1) - 0.9 = 1 - 3 - 0.9 = -2.9$$ 6. **Analyze signs:** Both $$f(0) = -0.9$$ and $$f(1) = -2.9$$ are negative. 7. **Conclusion:** Since $$f(0)$$ and $$f(1)$$ do not have opposite signs, the Intermediate Value Theorem does not guarantee a root in the interval $$(0,1)$$. **Final answer:** The Intermediate Value Theorem does not imply that the equation $$x^3 - 3x - 0.9 = 0$$ has a root in the interval $$(0,1)$$.