# Use the Mean Value Theorem to prove inequality

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!

• What does the MVT say if $y>x$ and $f(y)\leq f(x)$? May 21, 2015 at 3:07
• oh.. that I do not know.. May 21, 2015 at 3:12
• 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. May 21, 2015 at 3:19
• what do you mean the MVT version? do you mean the f(b)-f(a) all divided by b-a? May 21, 2015 at 3:23
• 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. May 21, 2015 at 3:28

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 $$y \gt x$$.