graph (1 Viewer)

[shkspeare]

can some1 help me sketch the graph y=(x²+1)/(e^x)?

i keep getting it wrong... the answers say there is two inflexions but i only get one inflexion pt at (1,2/e)

thxthxs

 

[Affinity]

there's one at (-1, 2e)

EDIT: bah.. my bad

 

[redslert]

errr no

graph generated from GrafEq 2.11

 

[CM_Tutor]

There are two inflexions: (1, 2 / e) is a horizontal POI, and (3, 10 / e³) is an ordinary POI. The horizontal POI is the only stationary point, at the curve is decreasing everywhere else.

 

[shkspeare]

how did u find the second POI =\

 

[CM_Tutor]

y = (x² + 1) / e^x

dy/dx = [e^x * (2x) - (x² + 1) * e^x] / (e^x)²
= e^x(2x - x² - 1) / e^2x
= (2x - x² - 1) / e^x
= -(x² - 2x + 1) / e^x
= -(x - 1)² / e^x

The only stationary point is located at x = 1, and everywhere else, dy/dx < 0, and so (1, 2 / e) is a horizontal POI.

d²y / dx² = [e^x * -2(x - 1)¹ * 1 - -(x - 1)² * e^x] / (e^x)²
= e^x(x - 1)[-2 + (x - 1)] / e^2x
= (x - 1)(x - 3) / e^x

So, possible inflexions at x = 1 and x = 3. Testing shows d²y/dx² changes sign around these values, so they are inflexions.

Note: It is clear from the graph that there must be two points of inflexion.