Question

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

Answer 1

Answer:

The domain is all real numbers where

$(f \circ g)(x)=\frac{x^2+6x+10}{9x^2+54x+88}$

Step-by-step explanation:

$(f \circ g)(x)=f(g(x))$

So g(x) must exist before plugging it into f(x).

Let's find where g(x) doesn't exist.

$x^2+6x+10$ is a quadratic expression.

$b^2-4ac$ is the discriminant and will tell us if $x^2+6x+10=0$ 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 $b^2-4ac$ we may not have to go full quadratic formula on this problem.

$b^2-4ac=(6)^2-4(1)(10)=36-40=-4$.

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.

$f(g(x))$

$f(\frac{1}{x^2+6x+10})$  

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

$\frac{1}{-2(\frac{1}{x^2+6x+10})+9}$  

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).

$\frac{1}{-2(\frac{1}{x^2+6x+10})+9} \cdot \frac{(x^2+6x+10)}{(x^2+6x+10)}$  

Using distributive property:

$\frac{1(x^2+6x+10)}{-2(\frac{1}{x^2+6x+10})\cdot(x^2+6x+10)+9(x^2+6x+10)}$

We are going to distribute a little more:

$\frac{x^2+6x+10}{-2+9x^2+54x+90}$

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

$\frac{x^2+6x+10}{9x^2+54x+88}$

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

$b^2-4ac=(54)^2-4(9)(88)=-252$

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

The domain is all real numbers where

$(f \circ g)(x)=\frac{x^2+6x+10}{9x^2+54x+88}$

Answer 2

Answer:

f (gx) = 1/ -2(1/x^2+6x+10) + 9

Step-by-step explanation:

f (gx) = 1/ -2(1/x^2+6x+10) + 9