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?
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)
Thank you so, so much!
To determine when the graph of a function is concave down, you need to find the intervals where the second derivative is negative. In this case, you correctly found the second derivative of the function to be 16e^(-x^2)(2x^2-1). Now, let's go through the steps to find the intervals where the second derivative is negative.
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).