curve sketching (1 Viewer)

Trebla

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how do u sketch y=e^[2x/(1+x^2)]
Sketch y = 2x / (1 + x2) (using division of ordinates or otherwise)

Note that:
- As 2x / (1 + x2) --> 0+, e2x / (1 + x2) --> 1+
- As 2x / (1 + x2) --> 0-, e2x / (1 + x2) --> 1-
- When 2x / (1 + x2) = 0 at the origin then e2x / (1 + x2) = 1
- The x-values of the turning points are the same (this is true for any composite function and is easily proven by differentiation)

From this you should be able to deduce the graph...
 

fullonoob

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how do u sketch y=e^[2x/(1+x^2)]
firstly graph y = 2x (linear function)
then graph 1+x^2 (parabola with vertex (0,1)
Through translation of graph do 2x / 1+x^2
then you have some graph looks like an oblique parabola thing (just guessing didnt actually do it)
now you graph y = e^x (just the basic graph going at y>0 and touches (0,1)

now since you have your graph y = e^x and 2x/(1+x^2)
simply use y =e^x and the other graph to translate it - plotting points in each plane or at turning points etc. :spin:
 

ronnknee

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firstly graph y = 2x (linear function)
then graph 1+x^2 (parabola with vertex (0,1)
Through translation of graph do 2x / 1+x^2
then you have some graph looks like an oblique parabola thing (just guessing didnt actually do it)
now you graph y = e^x (just the basic graph going at y>0 and touches (0,1)

now since you have your graph y = e^x and 2x/(1+x^2)
simply use y =e^x and the other graph to translate it - plotting points in each plane or at turning points etc. :spin:
What is "Through translation"?

The reply before this is the best method.
 

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