Hi everyone, I'm new here. I would appreciate it if someone can help me with this question. I need to find the range of the following function f(x) = (3x-2)^(1/2009). Thanks for any help
Let $\displaystyle g(x)=\sqrt[2009]{x}, \ h(x)=3x-2$.
Then $\displaystyle f(x)=\sqrt[2009]{3x-2}=g(h(x))$.
g and h are continuous and strictly increasing, so f is continuous and strictly increasing.
$\displaystyle \lim_{x\to-\infty}f(x)=-\infty, \ \lim_{x\to\infty}f(x)=\infty$.
Then the range of f is $\displaystyle \mathbb{R}$ (the set of real numbers).