Maths question (1 Viewer)

[Lucas_]

Hi, need help

Find the stationary point on the curve y = e^x and determine its nature. Find any points of inflexion and find values of y as x becomes very large or small. Hence sketch the curve.

I got as far as working out the stat point's co-ordinates and then what y'' is but now im stumped. Please help (barbernator)

 

[deswa1]

There's no stationary point on y=e^x- the curve is increasing for all values of x as the derivative is also e^x which is >0 for all x. Are you sure you wrote the question out properly?

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=y=xe^x\\ y'=e^x@plus;xe^x~(product~rule)\\ y''=2e^x@plus;xe^x~(product~rule)\\ intercepts~(0,0)\\ test~for~stat~points\\ e^x@plus;xe^x=0\\ x=-1\\ (-1,\frac{-1}{e})\\ test~for~concavity\\ at~x=-1\\ y''=\frac{1}{e}>0~(min~stat~point)\\ test~inflection~points\\ x=-2\\ (-2,\frac{2}{e^x})\\" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y=xe^x\\ y'=e^x+xe^x~(product~rule)\\ y''=2e^x+xe^x~(product~rule)\\ intercepts~(0,0)\\ test~for~stat~points\\ e^x+xe^x=0\\ x=-1\\ (-1,\frac{-1}{e})\\ test~for~concavity\\ at~x=-1\\ y''=\frac{1}{e}>0~(min~stat~point)\\ test~inflection~points\\ x=-2\\ (-2,\frac{2}{e^x})\\" title="y=xe^x\\ y'=e^x+xe^x~(product~rule)\\ y''=2e^x+xe^x~(product~rule)\\ intercepts~(0,0)\\ test~for~stat~points\\ e^x+xe^x=0\\ x=-1\\ (-1,\frac{-1}{e})\\ test~for~concavity\\ at~x=-1\\ y''=\frac{1}{e}>0~(min~stat~point)\\ test~inflection~points\\ x=-2\\ (-2,\frac{2}{e^x})\\" /></a>
as x-->infinity, y---> infinity.
x--->negative intinity,y--> 0-

so the graph is like this

 

[barbernator]

There's no stationary point on y=e^x- the curve is increasing for all values of x as the derivative is also e^x which is >0 for all x. Are you sure you wrote the question out properly?

he means xe^x

 

[Lucas_]

Barbenator wins , thanks