# How would you find the inflection point and the concavity of g(x) = (5x - 2.6) / (5x - 6.76)^2? I know I have to take the 2nd derivative but i'm not sure how because of the odd way this function is set up.?

Apr 24, 2015

Use the quotient rule to find $g ' \left(x\right)$, then simplify if possible, then use the quotient rule again to find $g ' ' \left(x\right)$.

$g \left(x\right) = \frac{5 x - 2.6}{5 x - 6.76} ^ 2$

$g ' \left(x\right) = \frac{5 {\left(5 x - 6.76\right)}^{2} - \left(5 x - 2.6\right) 2 \left(5 x - 6.76\right) \left(5\right)}{5 x - 6.76} ^ 4$

$= \frac{5 \left(5 x - 6.76\right) \left[\left(5 x - 6.76\right) - 2 \left(5 x - 2.6\right)\right]}{5 x - 6.76} ^ 4$

$= \frac{5 \left[5 x - 6.76 - 10 x + 5.2\right]}{5 x - 6.76} ^ 3$

$g ' \left(x\right) = \frac{- 5 \left(5 x + 1.56\right)}{5 x - 6.76} ^ 3$

Differentiate again to get $g ' ' \left(x\right)$.

After finding $g ' ' \left(x\right)$, proceed as in any other question about concavity.