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Question
find the equation of the line that is tangent to y=x^2-(5x^1/2)-10 at x=4
The equation of the line that is tangent to y=x^2-(5x^1/2)-10 is of the form : mx+n -> where m=f'(x) at x=4, & n will find it later.
Ok, y'(x)=2x-(5/2)x^(-1/2), y'(4)=8-(5/2)4^(-1/2)=7.2
Let's calculate y(4)=4^2-(5*4^1/2)-10=-4.
Our tangent line is 7.2x+n, let's find n :
-4=7.2*4+n --> n=-32.8
So, our tangent line is : 7.2x-32.8
Alon.
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Alon Mandes
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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 .
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3. 2 years of experience as a math teacher in college.
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5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
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M.A in Mathematics & Bs.c in Electronics.
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