6
$\begingroup$

Let $f(x)$ be continues function at $[1,9]$ and differentiable at $(1,9)$ and also $f(1) = 20 , f(9) = 68 $ and $ |f'(x)| \le 6$ for every $x \in (1,9)$.

I need to prove that $f(7) = 56$.

I started by using the Lagrange theorem and found that there exist $ 1<c<9$ such that $f'(c) = 6$ but I'm not sure how is this relevant and how to proceed.

$\endgroup$
  • $\begingroup$ Use the lagrange theorem twice, in proper intervals $\endgroup$ – Matheo Feb 6 '17 at 14:15
  • 1
    $\begingroup$ Hint: one candidate for $f(x)$ is just the linear polynomial interpolating those two points. Now, suppose that your $f(x)$ was less than this function for some value in the interval. $\endgroup$ – lulu Feb 6 '17 at 14:16
18
$\begingroup$
  1. Assume $f(7)>56$. What does the mean value theorem say exists somewhere in the interval $(1,7)$?
  2. Assume $f(7)<56$. What does the mean value theorem say exists somewhere in the interval $(7,9)$?

Conclusion: $f(7)\not>56$ and $f(7)\not<56$, so we must have $f(7)=56$.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

I would like to say that, with the given conditions, the only possible function is a line, connecting $(1,20)$ and $(9,68)$.

Let's say the curve goes off below the line, at some point $(x1,y1)$. Then, it has to return to the line eventually, as it needs to get to $(9,68)$, at some point $(x2,y2)$. The derivative of $f(x)$, in the interval $(x1,x2)$, should at some point become equal to $\frac{y2−y1}{x2−x1}>6$.

If it goes off, above the line, then we may apply a rather same reasoning.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But why is that the only possibility? $\endgroup$ – Arthur Feb 6 '17 at 14:23
  • 3
    $\begingroup$ Let's say the curve goes off below the line, at some point $(x_1,y_1)$. Then, it has to return to the line eventually, as it needs to get to $(9,68)$, at some point $(x_2,y_2)$. The derivative of $f(x)$, in the interval $(x_1,x_2)$ should at some point become equal to $\frac{y_2-y_1}{x_2-x_1}>6$. If it goes off above the line, then we may apply the same reasoning $\endgroup$ – Med Feb 6 '17 at 14:28
  • $\begingroup$ Good. That makes your answer into an actual answer, and not just hearsay. $\endgroup$ – Arthur Feb 6 '17 at 14:32
  • $\begingroup$ I should have given more detail. But, just wanted to give a way of thinking $\endgroup$ – Med Feb 6 '17 at 14:34
  • 1
    $\begingroup$ @DanielR.Collins, I took your advice. Thanks for caring. $\endgroup$ – Med Feb 6 '17 at 16:25
2
$\begingroup$

Hint: Consider $g(x)=f(x)-6x$, values and derivative, or alternatively $h(x)=6x+14-f(x)$

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.