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Question
find a point on the graph of y=e 5x at which the tangent line passes through the origin
Tangent line that passes through the origin will have the equation : y=mx . Where m is the
slope of the line %26 it's equal to the derivative of the function at point (xo,yo).
Let's calculate derivative :
f(x)=e^(5x)
f'(x)=5e^(5x)
f'(xo)=5e^(5xo) .
We also know that this line passes through the point (xo,yo) , therefore :
yo=mxo --> yo=5xoe^(5xo). We know that yo=5e^(5xo), therefore :
5e^(5xo)=5xoe^(5xo).
xo=1.
That means yo=5e^5 . So the point is (1,5e^5) .
Alon.
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Answers by Expert:
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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