help! photo attached

Question

anonymous · Answer

attachment

ghazi · Answer

is it 10.50 a.m there?

ghazi · Answer

$$f(x)=\frac{ 5t }{2+t^2  }, f'(x)=\frac{  (2+t^2)5+5t(2t) }{ (2+t^2) }$$ now you can do it :)

ghazi · Answer

sorry x=t i misplaced variable :P

anonymous · Answer

Haha its okay!

anonymous · Answer

so do i expand and solve it?

ghazi · Answer

$$\frac{ 15t^2+10}{ 2+t^2 }=0, 3t^2+2=0, t= \pm \frac{ 2 }{ 3 }$$

ghazi · Answer

now you have two values of t , differentiate your f'(t) once more

ghazi · Answer

the point where you will get value <0 you will have maxima so it could be either 2/3 or -2/3

ghazi · Answer

my calculation says at 2/3 you will have minimum and at -2/3 you will have maximum

anonymous · Answer

so value t is max and min ?

ghazi · Answer

no you have to find value by putting 2/3 in f(x) but at 2/3 you will have minimum of function and at -2/3 you will have maximum value, now to have the maximum value you have to put 2/3 and -2/3 in the function

ghazi · Answer

|dw:1363881788844:dw| lets say this is your graph so this graph is just an example to show you what i did, i found the points where function will have its maximum and minimum value now put 2/3 in f(t) you will have minimum value and same with -2/3 for max