Question

find the intervals on which the graph of f is concave upward, the intervals on which the graph of f is concave downward, and the inflection points.
$f(x)=\ln\left(x^{2}-6 x + 25\
ight)$
for what interval(s) of x is the graph of f concave upward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
$(-1,7)$
(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)
the graph is never concave upward.
for what interval(s) of x is the graph of f concave downward? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
$(-\infty,-1),(7,\infty)$
(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)
the graph is never concave downward.
determine the x coordinates of any inflection points of the graph of f(x). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a $x=$ (type an exact answer. use a comma to separate answers as needed.)
b. there are no inflection points.

Explanation:

Step1: Find the first derivative

Use the chain rule. If \(y = \ln(u)\) and \(u=x^{2}-6x + 25\), then \(y^\prime=\frac{u^\prime}{u}\).
Since \(u^\prime = 2x-6\), we have \(f^\prime(x)=\frac{2x - 6}{x^{2}-6x + 25}\).

Step2: Find the second derivative

Use the quotient rule \((\frac{v}{w})^\prime=\frac{v^\prime w - vw^\prime}{w^{2}}\), where \(v = 2x-6\) and \(w=x^{2}-6x + 25\).
\(v^\prime=2\) and \(w^\prime=2x - 6\).

$$
LATEXBLOCK0
$$

The denominator \((x^{2}-6x + 25)^{2}=( (x - 3)^{2}+16)^{2}>0\) for all \(x\in R\).
Set \(f^{\prime\prime}(x)>0\):
\(\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}}>0\), since \(- 2<0\) and \((x^{2}-6x + 25)^{2}>0\), then \((x + 1)(x - 7)<0\).
The solution of \((x + 1)(x - 7)<0\) is \(x\in(-1,7)\) (concave upward).
Set \(f^{\prime\prime}(x)<0\):
\(\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}}<0\), since \(-2<0\) and \((x^{2}-6x + 25)^{2}>0\), then \((x + 1)(x - 7)>0\).
The solution of \((x + 1)(x - 7)>0\) is \(x\in(-\infty,-1)\cup(7,\infty)\) (concave downward).
Set \(f^{\prime\prime}(x) = 0\):
\(\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}}=0\), then \(x=-1\) or \(x = 7\) (inflection points).

Answer:

• Concave upward: \((-1,7)\)
• Concave downward: \((-\infty,-1),(7,\infty)\)
• Inflection points: \(x=-1,x = 7\)