Question

suppose that 3 ≤ f(x) ≤ 4 for all values of x. what are the minimum and maximum possible values of f(8) - f(3)?
≤ f(8) - f(3) ≤

Explanation:

Step1: Apply Mean - Value Theorem

By the Mean - Value Theorem, if $y = f(x)$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then $f(b)-f(a)=f^{\prime}(c)(b - a)$ for some $c\in(a,b)$. Here, $a = 3$, $b = 8$, so $f(8)-f(3)=f^{\prime}(c)(8 - 3)=5f^{\prime}(c)$ where $c\in(3,8)$.

Step2: Find the minimum value

Since $3\leq f^{\prime}(x)\leq4$ for all $x$, when $f^{\prime}(c)=3$ (the minimum value of $f^{\prime}(x)$), $f(8)-f(3)=5f^{\prime}(c)$ has a minimum value. Substitute $f^{\prime}(c)=3$ into $5f^{\prime}(c)$, we get $5\times3 = 15$.

Step3: Find the maximum value

When $f^{\prime}(c)=4$ (the maximum value of $f^{\prime}(x)$), $f(8)-f(3)=5f^{\prime}(c)$ has a maximum value. Substitute $f^{\prime}(c)=4$ into $5f^{\prime}(c)$, we get $5\times4=20$.

Answer:

$15\leq f(8)-f(3)\leq20$