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Calculus/calculus help, please?

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Expert: Alon Mandes - 9/24/2008


Question
AP Calculus would be a lot more fun if i knew what the hell i was doing. Can you help me out of this dense fog?

find the horizontal tangents of the curve:
y=x^3-2x^2+x+1

Answer
The horizontal tangents are the lines that parallel to the x-axis.
That means that the slope of these lines is zero !!
So, we derive our function ,& we set it equal to zero & we find
these points. Let's do it :
y'=3x^2-4x+1 --> y'=0 --> 3x^2-4x+1=0 --> x1=1 & x2=-(1/3).
Jessica, these are the points on x-axis where the slope is zero.
But we also need to find in which point of y these lines pass through. To, know that , we simply can calculate y(x1) & y(x2) :
y(1)=1^3-2*1^2+1+1=2
y(-1/3)=(-1/3)^3-2(-1/3)^2-1/3+1=0.4
& these are the horizontal tangents lines of the curve .

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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