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)
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?
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)
First, find the critical points by setting the factor 2x^2-1 equal to zero:
2x^2 - 1 = 0
Solving this equation, we get:
2x^2 = 1
x^2 = 1/2
x = ±√(1/2) = ±(1/√2) = ±(√2/2)
So, the critical points are x = √2/2 and x = -√2/2.
Next, we determine the intervals by testing the sign of the second derivative in between and outside the critical points.
1. Pick a test point in the interval (-∞, -√2/2):
Choose x = -1. Substitute this into the second derivative:
f''(-1) = 16e^(-(-1)^2)(2(-1)^2-1)
= 16e^(-1)(2-1)
= 16e^(-1)
Since e^(-1) is positive, the whole expression is positive.
So, the second derivative is positive for x < -√2/2.
2. Pick a test point in the interval (-√2/2, √2/2):
Choose x = 0. Substitute this into the second derivative:
f''(0) = 16e^(-0^2)(2(0)^2-1)
= 16e^(0)(-1)
= -16
Since -16 is negative, the second derivative is negative for -√2/2 < x < √2/2.
3. Pick a test point in the interval (√2/2, ∞):
Choose x = 1. Substitute this into the second derivative:
f''(1) = 16e^(-1^2)(2(1)^2-1)
= 16e^(-1)(2-1)
= 16e^(-1)
Since e^(-1) is positive, the whole expression is positive.
So, the second derivative is positive for x > √2/2.
Based on the analysis above, we can conclude that the graph of y = 8e^(-x^2) is concave down on the interval (-√2/2, √2/2).
Expressing this answer in interval notation, we have: (-√2/2, √2/2).