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Expert: Alon Mandes - 8/17/2009


Question
The curve with equation y=(e^-x)(sinx) has one stationary point for which 0<=x<=pi. I) Find the x-coordinate of this point. II) Determine whether this point is a maximum or a minimum point. Thanks

Answer
Our curve is : y=[e^(-x)][sin(x)]. Let's derive using the rule : (fg)'=f'g+fg' :
y'=[e^(-x)]'[sin(x)]+[e^(-x)][sin(x)]'
y'=[-e^(-x)][sin(x)]+[e^(-x)][cos(x)]
y'=[e^(-x)][cos(x)-sin(x)] .

I) Stationary points or extremum point are yield when solving the equation y'=0. Therefore,
[e^(-x)][cos(x)-sin(x)]=0
Only 1 of 2 factors can be zero, only [cos(x)-1]. The factor [e^(-x)] can never be zero.Hence,
cos(x)=sin(x) --> x=pi/4 . This is the x-coordinate of the stationary point.

II) We need to chek if y''(x)>0 or y''(x)<0 .
If y''(x) < 0 then x is a maximum point .
If y''(x) > 0 then x is a minimum point .
Ok, lets derive 2nd time :
y''= { [e^(-x)][cos(x)-sin(x)] } '
y''=   [e^(-x)]'[cos(x)-sin(x)]+[e^(-x)][cos(x)-sin(x)]'
y''=   [-e^(-x)][cos(x)-sin(x)]+[e^(-x)][-sin(x)-cos(x)]
y''=   [e^(-x)][cos(x)+sin(x)]
Now,
y''(x=0) = [e^(0)][cos(0)+sin(0)] = 1*[1-0]=1 >0 It's a maximum point .

Alon.  

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Alon Mandes

Expertise

Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .

Experience

1. I'm a team member of mathnerds (math site for answering questions) 2. I'm a team member in the Student's Union of the Technion, helping students who have problems in mathematics. 3. 2 years of experience as a math teacher in college. 4. I give free homework help for high school students in Mathematics & Physics. 5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" , "Complex Functions".

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Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.

Education/Credentials
M.A in Mathematics & Bs.c in Electronics.

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