Expert: Alon Mandes - 12/29/2009

Question
QUESTION: can you please help me with those :
find the critical numbers for :
h(t)=t^(3/4)-2t^(1/4)
f(x)=x^(4/5)(x-4)^2
& find the extreme values for the function :
f(t)=2cos(t)+sin(2t) in the closed interval{0,PI/2}
I understand the concept of solving such problems; that we must get the dervative then equel it to zero but am having some difficultis in simplifying them so i will be really thankful if  show me step by step how to slove these three qusetion >


ANSWER: h(t)=t^(3/4)-2t^(1/4)
h'(t)=(3/4)t^(-1/4)-(1/2)t^(-3/4)
h'(t)=0 --> (3/4)t^(-1/4)=(1/2)t^(-3/4) --> (3/2)t^(-1/2) --> t=4/9 .
Now you have to calculate h(4/9) .

f(x)=[x^(4/5)][(x-4)^2]
f'(x)=[(4/5)x^(-1/5)][(-x4)^2]+2[x-4][x^(4/5)]
f'(x)=0 --> [(4/5)x^(-1/5)][(x-4)^2]=-2[x-4][x^(4/5)] -->
[(2/5)x^(-1/5)][(x-4)]=-[x^(4/5)] --> (2/5)(x-4)=-x --> (2/5)x-(8/5)+x=0 -->
(12/5)x=(8/5)
3x=2 --> x=(3/2) .
Now you have to calculate f(3/2) .

f(t)=2cos(t)+sin(2t) in the closed interval{0,PI/2}
f'(t)=-2sin(t)+2cos(2t)
f'(t)=0 --> sin(t)=cos(t) --> t=45 degrees  or t=pi/4 .
Now you have to calculate f(pi/4) .

Alon.





---------- FOLLOW-UP ----------

QUESTION: thanx alot for ur time , just to ,make sure in the second Question :
(2/5)x-(8/5)+x=0
x=8/7 not 3/2 ???
& in Question three :
sin(t)=cos(2t) not sin(t)=cos(t) is the answer the same anyway or is there something else to do >>>> thanks a lot anyway , u helped alot  

Answer
Yes. Sorry, my mistake , I apologize.

1. (2/5)x-(8/5)+x=0 --> x=8/7 . yes
2. sin(t)=cos(2t) --> We use the trigonometric identity : cos(2t0=1-2[sin(t)]^2 . So, we get
sin(t)=1-2[sin(t)]^2 . We set x=sin(t) to have : x=1-2x^2 --> 2x^2+x-1=0 , the solutions are:
x1=0.5  and  x2=-1 . Therefore :
x=0.5 --> sin(t)=0.5 --> t=30 degrees = pi/6
x=-1 --> sin(t)=-1 --> t=-90 degrees = -pi/2 "outside the range"
thus,
the solution is t=pi/6 .

Sorry again for the confusion .

Alon.