Math!

Yes Max, your algebra was fine in question 2 BUT your answer was not correct because you did not recognise the restriction on the domain.  

1.$\quad (f\cdot g)(x)\\=(x^2-7x+12)(\dfrac{7}{x^2}-16)\\ = 7 - \dfrac{49}{x}+\dfrac{84}{x^2}-16x^2+112x-192\\ =-16x^2+112x-185-\dfrac{49}{x}+\dfrac{84}{x^2}$

2.$f(x)=\sqrt{2x}-6\\ y=\sqrt{2x}-6\\ x = \sqrt{2y}-6\\ \sqrt{2y}=x+6\\ 2y = x^2 + 12x + 36\\ y = \dfrac{x^2}{2}+6x+18\\ f^{-1}(x)=\dfrac{x^2}{2}+6x+18$

1. Find f(g)(x) when f(x)=x^2 - 7x +12 and g(x)= 7/x^2 - 16.

idk Max, I think this is the answer...

$f(g(x)) \;\;or\;\; (f\cdot g)(x)\quad when\\ f(x)=x^2-7x+12 \quad and \quad g(x)=\frac{7}{x^2}-16\\~\\ \begin{align}  (f\cdot g)(x)&=(\frac{7}{x^2}-16)^2-7(\frac{7}{x^2}-16)+12\\ &=(\frac{49}{x^4}-[2*\frac{7}{x^2}*16]+256)-(\frac{49}{x^2}-112)+12\\ &=\frac{49}{x^4}-\frac{224}{x^2}+256-\frac{49}{x^2}+112+12\\ &=\frac{49}{x^4}-\frac{273}{x^2}+380\\ \end{align}$

2. Find the inverse function for f(x)=(sqrt2x-6).

When you do part 2 max you must be very careful of the domain.

f(x)=(sqrt2x-6).

$y=\sqrt{2x}-6$

Straight away I see that x must be greater or equal to 0 and y must be greater than or equal to -6

so when you take the inverse  x>=-6 and  y>=0

$f(x)=\sqrt{2x}-6\\ y=\sqrt{2x}-6\\ y+6=\sqrt{2x}\\ (y+6)^2=2x\\ x=\frac{(y+6)^2}{2}\\ \text{the inverse function becomes}\\ f^{-1}(x)=\frac{(x+6)^2}{2} \qquad where \quad x\ge-6$

Here are the 2 graphs drawn with Desmos graphing calculator.

NOTE that the invers of a function is the reflection of it across the line y=x.   It is important that you understand and remember this.

Luckily I am still correct in question 2, I just multiply out (x+6)^2 and divide all the terms by 2 to get rid of the fraction thing and made it ax^2 + bx + c.