State the point of inflection for the curve y = xe^-x
2007-06-26 09:09:24 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
my head hurts.
2007-06-26 09:11:38 · answer #1 · answered by Cheryl W 4 · 0⤊ 1⤋
No, the inflection point is where the second derivative is zero...
y = xe^(-x)
y' = e^(-x) - xe^(-x)
y'' = -e^(-x) - e^(-x) + xe^(-x)
y'' = e^(-x)*(-2+x)
this is zero at x=2 since e^(-x) is never zero (it can approach in the limit as x->inf, but is never EQUAL to zero...
inflection point at x=2.
2007-06-26 09:46:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Take first derivative and second derivative
y' = x(-e^-x) + e^-x(1) Chain Rule
y'' = x(e^-x) + -e^-x(1) + -e^-x
= x(e^-x ) - 2e^-x
= (x-2) (e^-x)
set equal to zero
x = 2
y = 2(e^-2) or y = .27 so point of inflection is (2,.27)
2007-06-26 09:25:51 · answer #3 · answered by gfulton57 4 · 0⤊ 0⤋
inflection point are where the slope is zero
`y= -xe^-x
let `y = 0
x = 0
2007-06-26 09:16:44 · answer #4 · answered by koki83 4 · 0⤊ 1⤋
According to http://en.wikipedia.org/wiki/Inflection_point
You will need to take the second derivative of your equation and find the point at which it changes sign (propably when the second derivative = 0 )
2007-06-26 09:14:53 · answer #5 · answered by rscanner 6 · 0⤊ 0⤋
It does not have a point of inflection since it is always concave downwards. It never changes its concavity.
2007-06-26 09:24:54 · answer #6 · answered by ironduke8159 7 · 0⤊ 1⤋
The secon derivative of your equation is:
f"(x)=(e^-x)-2e^-x
if you equate to zero and resolve
x=2
f"(x)=2/(e^2)
2007-06-26 09:27:25 · answer #7 · answered by ΛLΞX Q 5 · 0⤊ 0⤋
huh
2007-06-26 09:11:43 · answer #8 · answered by Git-r-done fan 3 · 0⤊ 1⤋
e=mc2
2007-06-26 09:12:46 · answer #9 · answered by philly_9112003 1 · 0⤊ 1⤋
make me!
2007-06-26 09:11:17 · answer #10 · answered by Sounds of Ed's football game 3 · 0⤊ 3⤋