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.
(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.

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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Set \(f^{\prime\prime}(x)=0\), then \(-2x^{2}+12x + 14 = 0\), or \(x^{2}-6x - 7=0\).
Factor: \((x - 7)(x+1)=0\), so \(x=-1\) or \(x = 7\).
The denominator \((x^{2}-6x + 25)^{2}>0\) for all real \(x\) (since \(x^{2}-6x + 25=(x - 3)^{2}+16>0\)).
Test intervals:

• For \(x<-1\), let \(x=-2\), \(f^{\prime\prime}(-2)=\frac{-2(-2)^{2}+12(-2)+14}{((-2)^{2}-6(-2)+25)^{2}}=\frac{-8-24 + 14}{(4 + 12+25)^{2}}=\frac{-18}{(41)^{2}}<0\).
• For \(-1
• For \(x>7\), let \(x = 8\), \(f^{\prime\prime}(8)=\frac{-2(8)^{2}+12(8)+14}{(8^{2}-6(8)+25)^{2}}=\frac{-128+96 + 14}{(64-48 + 25)^{2}}=\frac{-18}{(41)^{2}}<0\).

Answer:

A. \((-1,7)\)