SCalcET$9$

$4\mathrm{.}2\mathrm{.}029.$MI$.$SA$.$

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise

If $f(4)=2$ and $f^{\text{'}}(x)2$ for $4x7,$ how small can $f(7)$ possibly be$?$

Step $1$

Recall the mean value theorem, which states that if we let $f$ be a function that satisfies the following hypotheses, then there is a number $c$ in $(a,b)$ such that $f(b)-f(a)=f^{\text{'}}(c)(b-a).$

$f$ is continuous on the closed interval $a,b.$

$f$ is differentiable on the open interval $(a,b).$

Therefore, if the mean value theorem holds, we have the following for some $c$ in $(4,7).$

Step $2$

We are given the known value of the function $f(4)=2.$ Substituting allows us to find an expression for $f(7).$

$f(7)-f(4)=f^{\text{'}}(c)(7-4)$

$f(7)-2=3f^{\text{'}}(c)$

$f(7)=3f^{\text{'}}(c)2$

Additionally, we are given that $f^{\text{'}}(x)2$ for $4x7,$ so in particular we have $f^{\text{'}}(c)2.$ Putting $f^{\text{'}}(c)2$ into the equation allows us to find a minimal value for expression on the interval.

$f(7)=3r^{\text{'}}(c)23(,x)2$

$f(7)=3r^{\text{'}}(c)2,x$

To conclude, state how small f$(7)$ can possibly be$.$

$f(7)$

SKIP $($YOU CANNOT COME BACK$)$