Question

suppose that 3 ≤ f(x) ≤ 5 for all values of x. use the mean value theorem to find values for the inequality below. answer: □ ≤ f(6) - f(-2) ≤ □

Explanation:

Step1: Recall Mean - Value Theorem

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists a number \(c\in(a,b)\) such that \(f(b)-f(a)=f^{\prime}(c)(b - a)\). Here, \(a=-2\), \(b = 6\), so \(f(6)-f(-2)=f^{\prime}(c)(6-(-2))=8f^{\prime}(c)\) for some \(c\in(-2,6)\).

Step2: Use the given inequality for \(f^{\prime}(x)\)

We know that \(3\leq f^{\prime}(x)\leq5\) for all \(x\). Since \(c\in(-2,6)\), we substitute \(f^{\prime}(c)\) into the inequality. Multiply the inequality \(3\leq f^{\prime}(c)\leq5\) by \(8\). We get \(3\times8\leq8f^{\prime}(c)\leq5\times8\).

Step3: Simplify the inequality

\(24\leq8f^{\prime}(c)\leq40\). Since \(f(6)-f(-2)=8f^{\prime}(c)\), we have \(24\leq f(6)-f(-2)\leq40\).

Answer:

\(24\leq f(6)-f(-2)\leq40\)