the graph is concave down where
16e^-x^2 (2x^2-1) < 0
that is where
2x^2-1 < 0
x^2 < 1/2
so, f is concave down on the interval (-1/√2,1/√2)
For what values of x is the graph of
y = 8e^−x^2
concave down? (Enter your answer using interval notation.)
I started by finding the second derivative and factoring and ended up getting 16e^-x^2 (2x^2-1) and I know up till this part that I'm doing the right thing.
But then I had to find when each factor equaled zero and I got that 2x^2-1 = 0 at the sqrt of 1/2
and then I thought 16e^x^2 would equal 0 at 0 to infinity.
Then I set up a number line with points 0, sqrt 1/2 and infinity.
I found that between 0 and sqrt 1/2 the second derivative is negative and that between sqrt 1/ and infinity, the second derivative is positive.
So I tried (0, sqrt 1/2), [0, sqrt 1/2), and (0, sqrt 1/2] thinking that my interval notation was wrong, but I was still marked wrong.
How can I find the right answer?
2 answers
Thank you so, so much!