find the intervals on which the graph of ( f ) is concave upward, the intervals on which the graph of ( f )…

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.\n( f(x)=ln left(x^{2}-6 x + 25\right) )\nfor 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.\na.\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\nb. the graph is never concave upward.

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.\n( f(x)=ln left(x^{2}-6 x + 25\right) )\nfor 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.\na.\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\nb. the graph is never concave upward.

Answer

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). [ \begin{align*} f^{\prime\prime}(x)&=\frac{2(x^{2}-6x + 25)-(2x - 6)(2x - 6)}{(x^{2}-6x + 25)^{2}}\ &=\frac{2x^{2}-12x + 50-(4x^{2}-24x + 36)}{(x^{2}-6x + 25)^{2}}\ &=\frac{2x^{2}-12x + 50 - 4x^{2}+24x - 36}{(x^{2}-6x + 25)^{2}}\ &=\frac{-2x^{2}+12x + 14}{(x^{2}-6x + 25)^{2}} \end{align*} ] 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<x<7), let (x = 0), (f^{\prime\prime}(0)=\frac{-2(0)^{2}+12(0)+14}{(0^{2}-6(0)+25)^{2}}=\frac{14}{625}>0).
  • 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))