hi, when it says to sketch (f o g)(x) and $\displaystyle f(x) = x^2$ , $\displaystyle g(x) = 4x+3$ is that just $\displaystyle (4x+3)^2$ or $\displaystyle (4x+3)^2 \cdot x$ ? cheers
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Originally Posted by jvignacio hi, when it says to sketch (f o g)(x) and $\displaystyle f(x) = x^2$ , $\displaystyle g(x) = 4x+3$ is that just $\displaystyle (4x+3)^2$ or $\displaystyle (4x+3)^2 \cdot x$ ? cheers Its just the first one fog is a fuction like f When written as fog(x) its similar to writing f(x)
Originally Posted by ADARSH Its just the first one fog is a fuction like f When written as fog(x) its similar to writing f(x) ahh yeah so we dont multiply (f o g) by x, just leave it as (f o g) , in this case $\displaystyle (4x+3)^2$ correct?
Originally Posted by jvignacio ahh yeah so we dont multiply (f o g) by x, just leave it as (f o g) , in this case $\displaystyle (4x+3)^2$ correct? yes it is (fog)(x) = f(g(x))
Originally Posted by jvignacio ahh yeah so we dont multiply (f o g) by x, just leave it as (f o g) , in this case $\displaystyle (4x+3)^2$ correct? The notation f(x), with f a function, never means "multiply $\displaystyle f\cdot x$"!
Originally Posted by HallsofIvy The notation f(x), with f a function, never means "multiply $\displaystyle f\cdot x$"! yeah i know f(x) but i was confused with (f o g)(x) = f(g(x))
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