y= e^(2x)-e^(x)   sketch a curve by using derivative

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neela | High School Teacher | (Level 3) Valedictorian

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y=e^(2x)-e^x or

y=e^x(e^x-1).

y=0, when e^x=0  , when x approaches minus infinity on the left lower quadrant(3rd quadrant).

y=0, when e^x =1 or x = 0. So, the origin,(0 , 0) is a point n the curve.

y'=2e^(2x)-e^x= e^x(2e^x-1) =0. And y'' = 4e^(2x)-e^x.

Setting y' = 0 gives,  2e^x-1=0 for which x value is e^x=(1/2) or x = log(1/2). Also  y'(log(1/2)) = 0.5, wich is positive. Therefore, at x= log(1/2) , the function e^(2x)-e(x) has the minimum of (1/2)^2-1/2 = 0.25 at log(1/2) = -0.693(nearly).

As x--> minus infinity , y approaches to 0 from the minimum value of  -0.25  at x = -0.693 or log(1/2) .

Also , The area between x axis and the curve from -infinity to 0 is:  integral e^(2x-e^x)dx  between x=-inf to x = 0. is equal to:

[ e^2x/2-e^x] between x=-inf to x=0 .

=[((1/2)-1) - (0-0)] = -1/2..................(1)

Whereas,the area between 0 to 10 = [(1/2)e^20-e10]-(1/2-1) = 2.42456*10^8...............(2)

From (1) and (2) , we can see the curve approaches to 0 or wery narrow to x axis from x=0 to x=-infinity . And comaparatively it diveges suddenly  for  x > 0, covering large and larger area.

Thus y=0  or X axis is an asymtote to the curve.

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