Answer:
analyzed & sketched
Step-by-step explanation:
We are given the function:
[tex]y = \frac{\ln^2(x)}{9-x^2}[/tex]
The first derivative to find the region for increase and decrease:
[tex]y'= \frac{\frac{2\ln(x)}{x} (9-x^2)+2x\ln^2(x)}{(9-x^2)^2} =\frac{2\ln(x)(9-x^2+x^2\ln(x))}{x(9-x^2)^2} =0[/tex]
decreasing in (0,1), increasing in (1,3)∪(3,∞)
So, x = 1 is local minimum.
x = 3 is vertical asymptote.
The second derivative to find concave up and concave down:
[tex]y''=\frac{2\left(\left(x^2-9\right)^2+3x^2\left(x^2+3\right)\ln^2\left(x\right)+\left(-5x^4+54x^2-81\right)\ln\left(x\right)\right)}{x^2\left(9-x^2\right)^3}=0[/tex]
concave up in (0,3), concave down in (3,∞)
The sketch is given in the attachment.
