Number 28 is the only question I need please help, with steps

[tex]b^2-4ac[/tex] is the discriminant and will tell us if [tex]x^2+6x+10=0[/tex] will have any solutions. I'm trying to solve this equation because I want to figure out what to exclude from the domain. Depending on what [tex]b^2-4ac[/tex] we may not have to go full quadratic formula on this problem.

[tex]b^2-4ac=(6)^2-4(1)(10)=36-40=-4[/tex].

Since the discriminant is negative, then there are no real numbers that will make the denominator 0 here. So we have no real domain restrictions on g.

Let's go ahead and plug g into f.

[tex]f(g(x))[/tex]

[tex]f(\frac{1}{x^2+6x+10})[/tex]

I replaced g(x) with (1/(x^2+6x+10)).

[tex]\frac{1}{-2(\frac{1}{x^2+6x+10})+9}[/tex]

I replaced old input,x, in f with new input (1/(x^2+6x+10)).

Let's do some simplification now.

We do not like the mini-fraction inside the bigger fraction so we are going to multiply by any denominators contained within the mini-fractions.

I'm multiplying top and bottom by (x^2+6x+10).

[tex]\frac{1}{-2(\frac{1}{x^2+6x+10})+9} \cdot \frac{(x^2+6x+10)}{(x^2+6x+10)}[/tex]

Using distributive property:

[tex]\frac{1(x^2+6x+10)}{-2(\frac{1}{x^2+6x+10})\cdot(x^2+6x+10)+9(x^2+6x+10)}[/tex]

We are going to distribute a little more:

[tex]\frac{x^2+6x+10}{-2+9x^2+54x+90}[/tex]

Combine like terms on the bottom there (-2 and 90):

[tex]\frac{x^2+6x+10}{9x^2+54x+88}[/tex]

Now we can see if we have any domain restrictions here:

[tex]b^2-4ac=(54)^2-4(9)(88)=-252[/tex]

So again the bottom will never be zero because [tex]9x^2+54x+88=0[/tex] doesn't have any real solutions. We know this because the discriminant is negative.

The domain is all real numbers where

[tex](f \circ g)(x)=\frac{x^2+6x+10}{9x^2+54x+88}[/tex]