Why is f(x) = 1/2 (x^3) a vertical shrink?

Isn't is supposed to be a horizontal stretch, because f(x) = f(cx), 0<c<1?

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- Sep 21st 2013, 05:35 PMGragadoodleFunctional Transformations
Why is f(x) = 1/2 (x^3) a vertical shrink?

Isn't is supposed to be a horizontal stretch, because f(x) = f(cx), 0<c<1? - Sep 21st 2013, 05:59 PMvotanRe: Functional Transformations
horizontal shrink would be ((1/2)x)^3

- Sep 21st 2013, 06:02 PMGragadoodleRe: Functional Transformations
- Sep 22nd 2013, 01:55 PMFelixFelicis28Re: Functional Transformations
But then how do you make the distinction between $\displaystyle f\left(\tfrac{1}{2}x\right)$ and $\displaystyle \tfrac{1}{8} f(x)$ if $\displaystyle f(x) = x^3$? Or indeed, in OP's question, between $\displaystyle f\left(\tfrac{1}{2^{1/3}}x\right)$ and $\displaystyle \tfrac{1}{2} f(x)$?

OP, don't hold me to this, but I*think*the transformation acts as both but I'm not 100% sure.