5
$\begingroup$

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

  • 0
    Use the lagrange theorem twice, in proper intervals – 2017-02-06
  • 1
    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. – 2017-02-06

3 Answers 3

18
  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 $f(7)=56$.

3

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.

  • 0
    But _why_ is that the only possibility? – 2017-02-06
  • 3
    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 – 2017-02-06
  • 0
    Good. That makes your answer into an actual answer, and not just hearsay. – 2017-02-06
  • 0
    I should have given more detail. But, just wanted to give a way of thinking – 2017-02-06
  • 0
    Suggest that you edit your answer to include the upvoted comment above. – 2017-02-06
  • 1
    @DanielR.Collins, I took your advice. Thanks for caring. – 2017-02-06
2

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