Question

what are the inflection points f(x)=e^(-x^2)

Answer 1

Find its second derivative and make it equal to zero and solve for x

Answer 2

Look back at your previous question ames, those were your inflection points. :)
The x='s.

Answer 3

i could have sworn we just did this one

Answer 4

thats what i thought but then how do i tell if it concaved up or concaved down?

Answer 5

second derivative is
\[f''(x)=(4x^2-2)e^{-x^2}\] which is zero if
\[4x^2-2=0\] or
\[x=\pm\frac{1}{\sqrt2}\]

Answer 6

concave up if \(f''(x)>0\)

Answer 7

ignore the \(e^{-x^2}\) part as it is always posive
solve \(4x^2-2>0\)

Answer 8

\[y=4x^2-2\] is a parabola that faces up, so it is negative between the zeros and positive outside of them

Answer 9

i.e. \(f''(x)<0\) on the interval \((-\frac{1}{\sqrt2}, \frac{1}{\sqrt2})\)

Answer 10

positive otherwise
this more or less clear?

Answer 11

yes thank you.. i got a bit confused