Answer
(a) $f$ is decreasing on the interval $(-\infty,0)$
$f$ is increasing on the interval $(0,\infty)$
(b) The local minimum is $f(0) = 2.2$
(c) The graph is concave down on these intervals: $(-\infty,-3)\cup (3, \infty)$
The graph is concave up on this interval: $(-3,3)$
The points of inflection are $(-3,2.9)$ and $(3,2.9)$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = ln(x^2+9)$
We can find the points where $f'(x) = 0$:
$f'(x) = \frac{2x}{x^2+9} = 0$
$2x= 0$
$x = 0$
When $x \lt 0~~$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(-\infty,0)$
When $x \gt 0$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(0,\infty)$
(b) $f(0) = ln(0^2+9) = ln(9) = 2.2$
The local minimum is $f(0) = 2.2$
There is no local maximum.
(c) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{(2)(x^2+9)- (2x)(2x)}{(x^2+9)^2}$
$f''(x) = \frac{2x^2+18- 4x^2}{(x^2+9)^2}$
$f''(x) = \frac{18-2x^2}{(x^2+9)^2}$
$f''(x) = \frac{2(9-x^2)}{(x^2+9)^2} = 0$
$(2)(3-x)(3+x) = 0$
$x = -3, 3$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals: $(-\infty,-3)\cup (3, \infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-3,3)$
$f(-3) = ln[(-3)^2+9] = ln(18) = 2.9$
$f(3) = ln(3^2+9) = ln(18) = 2.9$
The points of inflection are $(-3,2.9)$ and $(3,2.9)$
(d) We can see a sketch of the graph below.