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Calculus1 10 Online
OpenStudy (anonymous):

h(f(g(x))) where f(x)=4x+3, g(x)=-5x, h(x)=6-2x

OpenStudy (anonymous):

h( f( g(x) ) ) = 6-2( 4 (-5x) + 3 )

OpenStudy (anonymous):

why is it 6-2( 4(-5x)+3) and not 6-2( 4x(-5x)+3)?

OpenStudy (anonymous):

Let's break this down step-by-step. I'm representing the functions only with their letter to clean it up a bit. \[=6-2*f(g)\] \[=6-2*(4(g)+3)\] \[=6-2*(4(-5x)+3)\] \[=6-8(-5x)+6\] \[=6+40x+6\] \[=40x+12\] The actual x variable only enters the equation after the final substitution because neither h nor f is a function of x. Only g is.

OpenStudy (anonymous):

aah! I get it now! Thank you so much!

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