Answer:
(6,1)
Step-by-step explanation:
[tex]f(x) = 5\sqrt[3]{x-6} + 1\\ f^{'}(x) = \frac{5}{3}(x-6)^{-\frac{2}{3}}\\f^{''}(x) = -\frac{10}{9}(x-6)^{-\frac{5}{3}}[/tex]
Now we set our 2nd derivative equal to 0 to ind the inflection point.
[tex]f^{''}(x) = -\frac{10}{9}(x-6)^{-\frac{5}{3}} = 0 => x = 6[/tex]
Now we plug in x = 6 into our original f(x).
f(6) = 1 => (6, 1)
