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Calculus/calculus 1 extrema of a function

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Expert: Alon Mandes - 11/4/2009


Question
locate the absolute extrema of the function y=xLn(x+3) on the interval of [0,3]

Answer
Lets derive y :
y'=Ln(x+3) + x/(x+3).
y'=0 --> Ln(x+3)=-x/(x+3) . There is no x in the interval [0,3] that satisfy the last equation.
Therefore ,the absolute extremums will be :
Absolute Minima , at x=0 --> y=0
Absolute Maxima , at x-3 --> y=3Ln(9)

Alon.

Calculus

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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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