Answer:
(6,1)
Step-by-step explanation:
To find the points of inflection, we need to find where the second derivative is equal to zero, or does not exist.
I will take the derivative using the chain rule.
[tex]f(x) = 5\sqrt[3]{(x-6) } + 1 \\\\f(x) = 5(x-6)^{\frac{1}{3} } + 1[/tex]
The first derivative:
[tex]f'(x) = 5*\frac{1}{3} (x-6)^{-2/3} \\\\f'(x) = \frac{5}{3} (x-6)^{-2/3} \\[/tex]
The second derivative:
[tex]f'(x) = \frac{5}{3} (x-6)^{-2/3} \\\\f''(x) = \frac{5}{3}*-\frac{2}{3} (x-6)^{-5/3}\\\\ f''(x) = -\frac{10}{9} (x-6)^{-5/3}[/tex]
Now to find the inflection points we have to find where the second derivative is equal to zero, or do not exist.
[tex]f''(x) = -\frac{10}{9} (x-6)^{-5/3}\\0= -\frac{10}{9} (x-6)^{-5/3}\\0= (x-6)^{-5/3}\\\\[/tex]
We can see that the second derivative does not exist when x=6, so there is an inflection point there.
We can solve the original equation to find the coordinate for x = 6.
f(x) = 5∛(x-6) + 1
f(6) = 5∛(6-6) + 1
f(6) = 1
So there is an inflection point at (6,1)
