Determine over what interval(s) is the function y=4x-6arctan(x) is concave up/concave down
I took the second derivative and ended up with x=0. I do not believe I did this right. IF I happen to be right, wouldn't our answer just be undefined, rather than having intervals being concave up or down?
If you got a single value for an interval, it doesn't look good for the kid.
y = 4x-6arctan(x)
y' = 4 - 6/(x^2+1) = (4x^2-1)/(x^2+1)
y" = 12x/(1+x^2)^2
You are correct with what you got. But, what do you have when y" = 0? An inflection point. That's where the concavity changes. So, now you have to remember that the sign of y" indicates concavity. Here,
y" < 0 for x<0
y" > 0 for x>0
Got it now? You can write your intervals.
Just to verify, check
http://www.wolframalpha.com/input/?i=+4x-6arctan%28x%29
To determine over what interval(s) the function is concave up or concave down, you need to find the intervals where the second derivative of the function is positive (concave up) or negative (concave down).
Let's start by finding the first and second derivatives of the given function:
First derivative:
dy/dx = 4 - 6/(1 + x^2)
Second derivative:
d^2y/dx^2 = 12x/(1 + x^2)^2
Now, to find the intervals where the second derivative is positive or negative, we set it equal to zero and solve for x:
d^2y/dx^2 = 12x/(1 + x^2)^2 = 0
Dividing both sides of the equation by 12x, we get:
x/(1 + x^2)^2 = 0
This equation is satisfied only when x = 0. However, when x = 0, the denominator (1 + x^2)^2 is not zero. Therefore, x = 0 is not a valid solution. This means that the second derivative does not equal zero for any x, and thus, it does not change sign.
Since the second derivative does not change sign, the function y = 4x - 6arctan(x) is not concave up or concave down over any interval(s). It is considered to be undefined in terms of concavity.
Therefore, your intuition is correct. There are no intervals where the function is concave up or concave down.