Question

What is the rate of increase for the function f(x) = One-third (RootIndex 3 StartRoot 24 EndRoot) Superscript 2 x?

Answer 1

Answer:

$4\sqrt[3]{9}$

Step-by-step explanation:

we have the function

$f(x)=\frac{1}{3}(\sqrt[3]{24})^{2x}$

Remember that

$24=(2^3)(3)$

substitute

$f(x)=\frac{1}{3}(\sqrt[3]{(2^3)(3)})^{2x}$

Applying property of exponents

$\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$

$(x^{m})^{n} =x^{m*n}$

so

$f(x)=\frac{1}{3}(\sqrt[3]{(2^3)(3)})^{2x}=\frac{1}{3}[(2^3)(3)}]^{\frac{2}{3}x}=\frac{1}{3}[(2^2)3^\frac{2}{3}]^x=\frac{1}{3}(4\sqrt[3]{9})^x$

therefore

The rate of increase is

$4\sqrt[3]{9}$

4RootIndex 3 StartRoot 9 EndRoot