Question

SOMEONE HELP PLEASE. What is the point of inflection of the following equation ( see reply ).

Answer 1

\[e ^{-2x}\]

Answer 2

Take the derivative twice and then set it to zero

Answer 3

i know.
could you help me do that?

Answer 4

dy/dx=-2e^(-2x)
take the derivative again: 4e^-2x

Answer 5

ok i got the second derivative i don't know how to solve for zero with that though.

Answer 6

2nd derivative = 4e^-2x. Set it equal to zero to find any values you can use to set up a second derivative test.

Answer 7

O wait I see what you mean..
so 4e^-2x=0
e^-2x=0
-2x=ln0
-2x=-infiniti
x=infiniti

Answer 8

There only exists an inflection point when the graph shifts from concave up and concave down.

Answer 9

so the graph doesn't shift from concave up or down?.. basically there is no point of inflection? because in my textbook there is an answer.

Answer 10

What is the answer in your text book.

Answer 11

in fact there are two points of inflection in my textbook.

Answer 12

lol i have no clue then :)

Answer 13

haha ok thanks anyways.

Answer 14

one of the points is (-1,root2)

Answer 15

i don't have the textbook with me though so i'm not 100 percent sure thats right.

Answer 16

if you have a graphing calculator on you, try to graph that equation. it might just be the case that you wrote the problem down wrong. I don't know though. Inflection points only occur if the graph shifts from concave up and concave down and vice versa. I cant seem to find any values to even attempt a second derivative test for inflection points so...

Answer 17

ok, thanks for your help.