Question 188982
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You have the right idea.  For *[tex \large \[g\,\circ\,h](x) = g\left(h(x)\right)] you substituted *[tex\large h(x)] where there was an <i>x</i> in *[tex\large g(x)].


For *[tex \large \[h\,\circ\,g](x) = h\left(g(x)\right)] you simply substitute *[tex \large g(x)] where there was an <i>x</i> in *[tex \large h(x)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  h(x) = x^2 + 3x + 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \[h\,\circ\,g](x)\ =\ h\left(g(x)\right)\ =\ (-2x)^2 + 3(-2x) + 2\ =\ 4x^2 - 6x + 2] 


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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