Question

assume that f(x) is continuous on the interval [2,6] and f inverse of x exists on the interval (2,6). if f(2)=1 and f prime x is less than or equal to 3/2 for all x in (2,6) show that f(6) is less then or equal to 7. ******i need a answer asap***** thanks!

Answer 1

i am not sure what the inverse has to do with this. this is like saying that the speed limit is
\[\frac{3}{2}\] and since you know at 2 you get 1, then how far can you be 4 units later

Answer 2

i meant f prime sorry!

Answer 3

if you had a line with slope
\[\frac{3}{2}\] and you knew that (2,1) was on the line you would also know that (6, 7) would be on the line as well, and that is as big as you can be. you want a proof using "mean value theorem"?

Answer 4

yea im not sure how to prove it

Answer 5

\[\frac{f(6)-f(2)}{6-2}=f'(c)\] for some c in the interval (2,6) but since you know
\[f'(x)\leq \frac{3}{2}\] for all x you must have
\[\frac{f(6)-f(2)}{6-2}\leq \frac{3}{2}\] so
\[\frac{f(6)-1}{4}\leq \frac{3}{2}\] therefore
\[f(6)-1\leq 6\] so
\[f(6)\leq 7\]