Question

1. - / 2 points details my notes ask your teacher suppose that ( 3 leq f^{prime}(x) leq 5 ) for all values of ( x ). what are the minimum and maximum possible values of ( f(8)-f(3) )? ( leq f(8)-f(3) leq ) resources read it master it

Explanation:

Step1: Apply the Mean Value Theorem

By the Mean Value Theorem, \(f(8)-f(3)=f^{\prime}(c)(8 - 3)=5f^{\prime}(c)\) for some \(c\in(3,8)\).

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

Since \(3\leq f^{\prime}(x)\leq5\), substitute \(f^{\prime}(c)\) into the inequality. Multiply each part of the inequality \(3\leq f^{\prime}(c)\leq5\) by \(5\).
We get \(3\times5\leq5f^{\prime}(c)\leq5\times5\).

Answer:

\(15\leq f(8)-f(3)\leq25\)