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

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.\n$(-1,7)$\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\nthe graph is never concave upward.\nfor 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.\n$(-\\infty,-1),(7,\\infty)$\n(type your answer in interval notation. type an exact answer. use a comma to separate answers as needed.)\nthe graph is never concave downward.\ndetermine 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.\na $x=$ (type an exact answer. use a comma to separate answers as needed.)\nb. there are no inflection points.
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}). 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). [ \begin{align*} f^{\prime\prime}(x)&=\frac{2(x^{2}-6x + 25)-(2x - 6)^{2}}{(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}}\ &=\frac{-2(x^{2}-6x - 7)}{(x^{2}-6x + 25)^{2}}\ &=\frac{-2(x + 1)(x - 7)}{(x^{2}-6x + 25)^{2}} \end{align*} ] 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)