(a) Find the intervals of increase or decrease. $\\$
(b) Find the local maximum and minimum values.$\\$
(c) Find the intervals of concavity and the inflection points.$\\$
(d) Use the information from parts $ (a) - (c) $ to sketch the graph. $\\$
Check your work with a graphing device if you have one.
$ f(x) = \ln (x^2 + 9) $
a)$f(x)$ is increasing on $x \in(0, \infty)$$\\$
$f(x)$ is decreasing on $x \in(-\infty, 0)$$\\$
b) Using first derivative test we can say that:$\\$
There is local minima at $x=0$$\\$
There is no local maxima$\\$
c) concave up: $(-3,3)$ concave dawn: $(-\infty,-3) \cup(3, \infty)$ inflection pts at $x=-3$ and 3$\\$
d) SEE GRAPH
derivatives and second derivatives are extremely useful because they allow us to determine the shape of graphs, increase decrease minimums and maximums and con cavity. So, uh, this graph is going to be the natural log of X squared plus nine. I think so. Looking at this graph here, we first want to determine intervals of increase and decrease. We see the graph is decreasing from negative infinity to zero, and then it increases from zero to infinity. Then we see that there's a local minimum right here at 2.92 point 197 That is the natural log of nine. And then that is the only local minimum or maximum. And we see this because if we take the derivative of the graph, we see that there's only one critical point. Then we can take the second derivative of the graph, and we see that there is, uh, there are two inflection points. So the graph is con cave down. Um, mostly until it reaches this point right here. Negative three. And then it goes. Khan gave up and then positive three. It goes con cave down again. Eso With that, we see that negative three and positive three are inflection points. And with all this information, we can combine all these pieces to construct this graph.