Question

I have an exam tomorrow!! please help me!!
If f''(x) exists for all x anf f(1)=5 f(3)=-2 f(5)=1, show that there are points c,d in (1,5) such that f'(c)=0 and f'(d)>5/4

Answer 1

you will need this ..
http://en.wikipedia.org/wiki/Mean_value_theorem

Answer 2

the fact that f(x) decreases from 1 to 3 then increases from 3 to 5 implies there was a turning point or a local minimum. The slope at this point must be 0. therefore there is some value c such that f'(c) = 0

Look at interval from 3 to 5. the slope between these 2 points is 3/2
--> m= (1-(-2))/(5-3) = 3/2
The mean value thm says there must be some value d within given interval such f'(d) = m.
since 3/2 > 5/4, thus there is some value d such that f'(d) > 5/4