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f is a continuous function defined on [a,b] and differentiable on ]a,b[ with f'(x)>0 on ]a,b[.

Use the mean value theorem to prove that for any x, y, all real [a,b], if y > x then f(y)>f(x)

I understand what the MVT is and what the collalories are, but I just can't figure out how to do these types of questions!

Thanks!

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  • $\begingroup$ What does the MVT say if $y>x$ and $f(y)\leq f(x)$? $\endgroup$
    – David K
    May 21, 2015 at 3:07
  • $\begingroup$ oh.. that I do not know.. $\endgroup$
    – user156248
    May 21, 2015 at 3:12
  • $\begingroup$ I see now there seems to be a missing fact in the problem statement. You must have $a\leq x < y \leq b$, otherwise you really can't say anything about the relationship of $f(x)$ and $f(y)$. But supposing $a\leq x < y \leq b$, then you have a function continuous on $[x,y]$ and differentiable on $]x,y[$, so the MVT says something about $f(x)$, $f(y)$, and $f'$. What does it say? If you cannot answer, then at least type the complete MVT into your question so we can at least know what version you are working with. $\endgroup$
    – David K
    May 21, 2015 at 3:19
  • $\begingroup$ what do you mean the MVT version? do you mean the f(b)-f(a) all divided by b-a? $\endgroup$
    – user156248
    May 21, 2015 at 3:23
  • $\begingroup$ Yes, that's part of the idea. As you can see in the answer, the $a$ and $b$ in the statement of the MVT don't always have to be $a$ and $b$. You can change all the $a$s to $x$s, for example. $\endgroup$
    – David K
    May 21, 2015 at 3:28

2 Answers 2

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From the MVT, if $y \gt x$, there exists some $c$ in interval $(x,y)$ such that slope of the chord joining $x$ and $y$ equals $f'(c)$. So, $\frac{f(y)-f(x)}{y-x}$ should equal $f'(c)$. Since it is given that the derivative is positive throughout the interval, $f'(c) \gt 0$. It should follow that $f(y)-f(x)$ is positive since $y-x$ is positive. Thus, $f(y) \gt f(x)$ if $y \gt x$.

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  • $\begingroup$ how would you answer this then: given g(x)= x - arctanx, prove that g'(x)>0 for x>0? $\endgroup$
    – user156248
    May 21, 2015 at 3:27
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Since f'(x) >0, the function is increasing so for every y > x in the interval the intercept on y axis f(y) would be greater than the previous one on the left side i.e f(x)

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