Answer: C. (0.73, 10)
Step-by-step explanation:
Maximum growth occurs at f'(x). Maximum growth rate occurs when f''(x) = 0
[tex]f(x) = \dfrac{20}{1+9e^{-3x}}[/tex]
[tex]f'(x) = \dfrac{540e^-3x}{(1+9e^{-3x})^2}[/tex]
[tex]f''(x)=\dfrac{1620e^{-6x}(-e^{3x}+9)}{(1+9e^{-3x})^3}[/tex]
[tex]0=\dfrac{1620e^{-6x}(-e^{3x}+9)}{(1+9e^{-3x})^3}[/tex]
[tex]0 = (1620e^{-6x})(-e^{3x}+9)[/tex]
[tex]1620e^{-6x}=0\\ e^{-6x}=0\\ ln\ e^{-6x}=ln\ 0\\ ln\ 0\text{\ is NOT VALID}[/tex]
[tex]-e^{3x}+9=0\\ 9=e^{3x}\\ ln\ 9=ln\ e^{3x}\\ ln\ 9=3x\\ \\ \dfrac{ln\ 9}{3}=x\\ \\ 0.73 = x[/tex]
[tex]f(0.73) = \dfrac{20}{1+9e^{-3(0.73)}}[/tex]
= 10
