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About 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".

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

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

You are here: Experts > Teens > Homework/Study Tips > Calculus > curve sketching

Calculus - curve sketching

Expert: Alon Mandes - 6/25/2009

Question
I need help with this problem.

Given the following function, state the domain, all asymptotes, intercepts, relative extrema, and inflection points. Then graph the function.

h(x)=(4-x^2)/(x^2+1)

Answer

graph of the curve

h(x)=(4-x^2)/(x^2+1)

1)Domain : All x ("because x^2+1 is always positive")

2)Intercepts with x-axis : y=0 -> (4-x^2)/(x^2+1) =0 -> (4-x^2)=0
  -> x=±2 .

3)Intercepts with y-axis : x=0 -> h(0)=(4-0)/(0+1)=4 -> y=4 .

4)Horizontal asymptotes :
Lim   (4-x^2)/(x^2+1) =
x->∞
Lim   (4/x^2 -1)/(1/x^2 +1) = (0-1)/(0+1) = -1
x->∞
y=-1 is the Horizontal asymptotes .

4)Vertical asymptotes : NONE. Because x^2+1 can never be zero !

Extremum points :
h'(x)=[ (-2x)(x^2+1)-(2x)(4-x^2) ] /(x^2+1)^2
h'(x)=[-2x^3-2x-8x+2x^3]/(x^2+1)^2
h'(x)=-10x/(x^2+1)^2
h'(x)=0 -> x=0 is a critical point !
h''(x)=[-10(x^2+1)^2-2x(-10x)]/(x^2+1)^4
h''(0)=-10
h''(0) < 0 then it's a minimum point.

The graph is attached below.

Alon.

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