SOLUTION: Find the intervals on which f is increasing and the intervals on which it is decreasing. f(x)= x^2 ln x^2+3

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Question 1155431: Find the intervals on which f is increasing and the intervals on which it is decreasing.
f(x)= x^2 ln x^2+3
Found 3 solutions by greenestamps, MathLover1, ikleyn:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Clarify....

x^2 ln(x^2) + 3 = x%5E2ln%28%28x%5E2%29%29%2B3?

or

x^2 ln(x^2+3) = x%5E2ln%28%28x%5E2%2B3%29%29?

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

f%28x%29=+x%5E2%2A+ln+%28x%5E2%2B3%29-> parabola, opens up
domain:
(-infinity,infinity)
range:
{ f element R : f%3E=0} (all non-negative real numbers)
x-intercept:
0=+x%5E2%2Aln%28x%5E2%2B3%29
x=0
=>x-intercept at origin

y-intercept:
y=+0%5E2%2Aln%280%5E2%2B3%29
y=0
=>y-intercept at origin

=> minimum is at origin
so, the interval on which f is increasing is
(0,infinity)

and the interval on which it is decreasing is
(-infinity,0)

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2A+ln+%28x%5E2%2B3%29%29+

Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.


    

    Plot y = (x^2)*(ln(x^2+3)



Looking into the formula, you can see that the function  f(x) = (x^2)*ln(x^2+3)  


    - first, is defined at all values of x (over all the domain of real numbers) and is even function,

    - and second, that it is MONOTONIC in the domain x >= 0.


Indeed, than larger the argument x is, than larger each of both factors  x^2  and ln(x^2+3) is.


So the function f(x) is monotonically increasing in the domain x >= 0.


Then from the fact that it is even function, you may conclude that the function is monotonically DECREASING in the domain  x < 0.


So you can perform all the necessary analysis without using Calculus, i.e., practically, MENTALLY.