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

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

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)^{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*} ] 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))