determine the open intervals on which the graph of the function is concave upward or concave downward…

Explanation:

Step1: Find the first derivative

Use the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here (u = x - 3), (u^\prime=1), (v = 7x + 6), (v^\prime = 7). [ \begin{align*} f^\prime(x)&=\frac{1\times(7x + 6)-(x - 3)\times7}{(7x + 6)^{2}}\ &=\frac{7x+6-(7x-21)}{(7x + 6)^{2}}\ &=\frac{7x+6 - 7x + 21}{(7x + 6)^{2}}\ &=\frac{27}{(7x + 6)^{2}} \end{align*} ]

Step2: Find the second derivative

Use the quotient rule again. Let (u = 27), (u^\prime=0), (v=(7x + 6)^{2}), (v^\prime = 2\times7\times(7x + 6)=14(7x + 6)) [ \begin{align*} f^{\prime\prime}(x)&=\frac{0\times(7x + 6)^{2}-27\times14(7x + 6)}{(7x + 6)^{4}}\ &=\frac{-378(7x + 6)}{(7x + 6)^{4}}\ &=\frac{-378}{(7x + 6)^{3}} \end{align*} ]

Step3: Find the domain and critical points

The function (y = f(x)) is undefined when (7x+6 = 0), i.e., (x=-\frac{6}{7}). The domain of (f(x)) is ((-\infty,-\frac{6}{7})\cup(-\frac{6}{7},\infty)). Set (f^{\prime\prime}(x)=0), but (\frac{-378}{(7x + 6)^{3}} = 0) has no solution.

Step4: Test the intervals

• For the interval ((-\infty,-\frac{6}{7})), let (x=-1). Then (f^{\prime\prime}(-1)=\frac{-378}{(-7 + 6)^{3}}=\frac{-378}{-1}=378>0)
• For the interval ((-\frac{6}{7},\infty)), let (x = 0). Then (f^{\prime\prime}(0)=\frac{-378}{(0 + 6)^{3}}=\frac{-378}{216}<0)

Answer:

concave upward: (\left(-\infty,-\frac{6}{7}\right)) concave downward: (\left(-\frac{6}{7},\infty\right))