I have a function xln(4-x^2) for -2

Question

I have a function xln(4-x^2) for -2<x<2
I also have the graph.
It asks me to get the values of x where the derivative is equal to 0...which seems imposible to get

ganeshie8 · Answer

uhm why impossible ?

osanseviero · Answer

I already got the derivative

$$\frac{ -2x^{2} }{ x^{2}-4 }+\ln (4-x ^{2})$$

osanseviero · Answer

Seems hard to get the values where the derivative intersects x

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=%282x%5E2%29%2F%28x%5E2-4%29+%2B+ln%284-x%5E2%29+%3D+0+++%2C++-2+%3C+x+%3C+2

ganeshie8 · Answer

yep it does look evil

osanseviero · Answer

Any other way to solve this?
It says "P and Q are points of the function where the tangent is parallel to the x-axis"

I thought of getting the derivative and saying that's equal to 0

ganeshie8 · Answer

i see nothing wrong in ur method...

osanseviero · Answer

Then it is an evil problem of ib?

ganeshie8 · Answer

one sec il give it a try

osanseviero · Answer

okk

ganeshie8 · Answer

yeahh solving it is not pleasant lol... id just wolfram it and say :
the tangent is parallel to x-axis at :  x =  -1.149, 1.149

ganeshie8 · Answer

unless you're taught other approximation methods...

osanseviero · Answer

Not yet...

osanseviero · Answer

Well...thanks for the help!

ganeshie8 · Answer

np :)