Question 708367
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If *[tex \LARGE \phi] is a function defined by *[tex \LARGE \phi(x)\ =\ x^2\ +\ 2x\ -\ 3], then if you are asked to evaluate *[tex \LARGE \phi] at some number, say, 2, which is to say evaluate *[tex \LARGE \phi(2)], you simply have to substitute 2 for every instance of *[tex \LARGE x] in the definition of *[tex \LARGE \phi(x)].  That is:  *[tex \LARGE \phi(2)\ =\ (2)^2\ +\ 2(2)\ -\ 3\ = 5]


Likewise, *[tex \LARGE \phi(a)\ =\ a^2\ +\ 2a\ -3] and *[tex \LARGE \phi(\chi\ -\ \gamma)\ =\ (\chi\ -\ \gamma)^2\ +\ 2(\chi\ -\ \gamma)\ -\ 3]


Doing a composite, *[tex \LARGE (\phi\ \circ\ \varphi)] is nothing more than evaluating the first function at the value of the second function, that is *[tex \LARGE \phi\left((\varphi(x)\right)\ =\ \left(\varphi(x)\right)^2\ +\ 2\left(\varphi(x)\right)\ -\ 3]


For your problem, you want *[tex \LARGE g] composite *[tex \LARGE f], which is just *[tex \LARGE g] of *[tex \LARGE f] of *[tex \LARGE x], or *[tex \LARGE g\left(f(x)\right)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ h(x)\ =\ \left(g\circ f\right)(x)\ =\ g\left(f(x)\right)\ =\ \left(x\ +\ 5\right)^2]


Whether or not your instructor expects you to expand the squared binomial is a matter between s/he and you.  In any case, you still need to evaluate *[tex \LARGE h(4)]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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