Function of a Function - Engineering Student - HELP!!!

[LukeGernon]

Struggling guys with these functions ive done loads and loads of simple/mediocre type questions and find them easy as, but ive hit questions like this and don't know where to even begin. Mainly with the (INEQUALITY) Symbol have never used it in practice before until now. please help guys.

[Baltuilhe]

Good night!

[tex]g(x)=\frac{6}{3-x}[/tex]
[tex]h(x)=5x+2[/tex]

a)
[tex]g(h(x))=\frac{6}{3-h(x)}=\frac{6}{3-(5x+2)}=\frac{6}{1-5x}[/tex]

b)
[tex]h(g(x))=5g(x)+2=5\cdot\frac{6}{3-x}+2=\frac{30}{3-x}+2=\frac{30+2(3-x)}{3-x}=\frac{36-2x}{3-x}[/tex]

Hope to have helped!

[HallsofIvy]

g(f(x)) means "replace every x in the definition of g(x) by f(x)".

Here [tex]g(x)= \frac{6}{3- x}[/tex] and [tex]f(x)= 5x+ 2[/tex] so [tex]g(f(x))=\frac{6}{3- (5x- 2)}= \frac{6}{1- 5x}[/tex].

f(g(x)) means "replace the x in f(x) by g(x)" so [tex]f(g(x))= 5\left(\frac{6}{3- x}\right)+ 2[/tex][tex]= \frac{30}{3- x}+ \frac{6- 2x}{3- x}[/tex][tex]= \frac{36- 2x}{3- x}[/tex].

(All my fractions show with no horizontal line. Is that just my viewer or do others see it that way?)

[HallsofIvy]

Although I posted this months ago, I should have included
[tex]g(f(x))= \frac{6}{1- 5x}[/tex] for [tex]x\ne \frac{1}{5}[/tex]

and
[tex]f(g(x))= \frac{36- 2x}{3- x}[/tex] for [tex]x\ne 3[/tex].