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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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?
horizontal shrink would be ((1/2)x)^3
Thank you so much!Quote:
Originally Posted by votan
I spent like 5 hours trying to figure this out!!!!!!1
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 $\displaystyle \tfrac{1}{2} f(x)$?Quote:
Originally Posted by votan
OP, don't hold me to this, but I think the transformation acts as both but I'm not 100% sure.
