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.
a ( (-1,7) )
(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)
b 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.
a
(type your answer in interval notation. type an exact answer use a comma to separate answers as needed.)
b. the graph is never concave downward

Explanation:

Step1: Find the first - derivative

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

Step2: Find the second - derivative

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

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Since \(x^{2}-6x + 25=(x - 3)^{2}+16>0\) for all \(x\in R\).
Set \(f^{\prime\prime}(x)>0\) (concave upward):
\(\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}}>0\), multiply both sides by \(-\frac{(x^{2}-6x + 25)^{2}}{2}\) (inequality sign flips), we get \((x + 1)(x - 7)<0\).
The solution of \((x + 1)(x - 7)<0\) is \(x\in(-1,7)\).
Set \(f^{\prime\prime}(x)<0\) (concave downward):
\(\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}}<0\), multiply both sides by \(-\frac{(x^{2}-6x + 25)^{2}}{2}\) (inequality sign flips), we get \((x + 1)(x - 7)>0\).
The solution of \((x + 1)(x - 7)>0\) is \(x\in(-\infty,-1)\cup(7,\infty)\).

Answer:

For concave upward: A. \((-1,7)\)
For concave downward: A. \((-\infty,-1)\cup(7,\infty)\)