4. evaluate the given function at x = 1 and x...

# Answer: We first evaluate \(f(1)\) and \(f(3)\) to check the Intermediate - Value Theorem conditions. 1. Evaluate \(f(1)\): - Substitute \(x = 1\) into \(f(x)=x^{3}-5x^{2}+3x + 6\). - \(f(1)=1^{3}-5\times1^{2}+3\times1 + 6\). - \(f(1)=1-5 + 3+6\). - \(f(1)=5\). 2. Evaluate \(f(3)\): - Substitute \(x = 3\) into \(f(x)=x^{3}-5x^{2}+3x + 6\). - \(f(3)=3^{3}-5\times3^{2}+3\times3 + 6\). - \(f(3)=27-45 + 9+6\). - \(f(3)=-3\). Since \(f(x)\) is a polynomial, it is continuous everywhere. And \(f(1)=5>0\) and \(f(3)=-3<0\). By the Intermediate - Value Theorem, which states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\), and \(k\) is a number between \(f(a)\) and \(f(b)\), then there exists at least one number \(c\) in the open interval \((a,b)\) such that \(f(c)=k\). In the case of \(k = 0\), since \(0\) is between \(f(1)\) and \(f(3)\), there is at least one zero of the function \(f(x)\) in the interval \((1,3)\). # Explanation: ## Step1: Evaluate \(f(1)\) Substitute \(x = 1\) into \(f(x)\): \(f(1)=1 - 5+3 + 6=5\) ## Step2: Evaluate \(f(3)\) Substitute \(x = 3\) into \(f(x)\): \(f(3)=27-45 + 9+6=-3\) ## Step3: Apply Intermediate - Value Theorem Since \(f(x)\) is continuous (as it's a polynomial) and \(f(1)>0,f(3)<0\), there is a zero in \((1,3)\)