Answer:
[tex]4\sqrt[3]{9}[/tex]
Step-by-step explanation:
we have the function
[tex]f(x)=\frac{1}{3}(\sqrt[3]{24})^{2x}[/tex]
Remember that
[tex]24=(2^3)(3)[/tex]
substitute
[tex]f(x)=\frac{1}{3}(\sqrt[3]{(2^3)(3)})^{2x}[/tex]
Applying property of exponents
[tex]\sqrt[n]{x^{m}}=x^{\frac{m}{n}}[/tex]
[tex](x^{m})^{n} =x^{m*n}[/tex]
so
[tex]f(x)=\frac{1}{3}(\sqrt[3]{(2^3)(3)})^{2x}=\frac{1}{3}[(2^3)(3)}]^{\frac{2}{3}x}=\frac{1}{3}[(2^2)3^\frac{2}{3}]^x=\frac{1}{3}(4\sqrt[3]{9})^x[/tex]
therefore
The rate of increase is
[tex]4\sqrt[3]{9}[/tex]
4RootIndex 3 StartRoot 9 EndRoot
