Let f(x) be a function such that f′(x)=x(ln|x|)^2.
Find the open interval(s) in which f(x) is concave down .
In interval notation, f(x) is concave down in ___________
f' = x (lnx)^2
f" = lnx(lnx+2)
f is concave down where f" < 0, or
-2 < lnx < 0
That is, on the interval (1/e^2,1)
You can see the graph of f(x) at
http://www.wolframalpha.com/input/?i=1%2F4+x^2+%282%28lnx%29^2-2lnx%2B1%29+for+0%3Cx%3C2
To find the open interval(s) in which f(x) is concave down, we need to determine the sign of the second derivative of the function f(x).
The given information about f(x) is that f′(x) = x(ln|x|)^2. To find the second derivative, we need to differentiate the first derivative with respect to x.
Derivative of f′(x) = d/dx[x(ln|x|)^2]
Using the product rule, we have:
f′′(x) = d/dx[x] * (ln|x|)^2
+ x * d/dx[(ln|x|)^2]
The first term (d/dx[x]) will give us 1. Now, let's differentiate the second term, x * d/dx[(ln|x|)^2] using the chain rule.
d/dx[(ln|x|)^2] = 2(ln|x|)(1/x) (Differentiating (ln|x|)^2 using the chain rule)
Simplifying this expression, we get:
f′′(x) = 1 + 2(ln|x|)(1/x)
To determine the concavity, we need to find the sign of f′′(x). Let's analyze the values of f′′(x) in different intervals.
For x > 0, we have:
f′′(x) = 1 + 2(ln|x|)(1/x)
= 1 + 2(lnx)(1/x) (Since |x| = x for x > 0)
= 1 + 2lnx/x
For x < 0, we have:
f′′(x) = 1 + 2(ln|x|)(1/x)
= 1 + 2ln(-x)/(-x) (Since |x| = -x for x < 0)
= 1 + 2ln(-x)/x
Now, let's analyze the sign of f′′(x) in the two cases:
1. For x > 0:
If we take a positive value of x (e.g., x = 1), f′′(x) = 1 + 2lnx/x > 0. This means in the interval (0, ∞), f(x) is concave up.
2. For x < 0:
If we take a negative value of x (e.g., x = -1), f′′(x) = 1 + 2ln(-x)/x > 0. This means in the interval (-∞, 0), f(x) is concave up as well.
Since f(x) is concave up in both the intervals (0, ∞) and (-∞, 0), there are no open intervals in which f(x) is concave down.
Therefore, in interval notation, f(x) is concave down in the empty set or "{}."