1. $$ f(x)=x^{2}+6 x+5 $$ and $$ g(x)=9-2 x $$ for all values of $$ x $$. (a) Find $$ g(2) $$ $$ \begin{aligned} & 9-2 \times 2 \ = & 5 \end{aligned} $$ (b) Find $$ g g(2) $$ (c) Find $$ f g(x) $$ in it simplest form. $$ \begin{aligned} & \left(x^{2}+6 x+5\right)(9-2 x) \ = & 9 x^{2}-2 x^{3}+54 x-12 x^{2}+45-10 x \ = & -2 x^{3}-3 x^{2}+44 x+45 \end{aligned} $$ (d) Find $$ f^{-1} $$ (e) Solve the equation $$ f(x)=2 x+17 $$

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Let's solve the problem step by step.

### (a) Find $$ g(2) $$

We have the function $$ g(x) = 9 - 2x $$. To find $$ g(2) $$:

$$
g(2) = 9 - 2 \times 2 = 9 - 4 = 5
$$

So, $$ g(2) = 5 $$.

### (b) Find $$ g(g(2)) $$

Since we found $$ g(2) = 5 $$, we now need to find $$ g(5) $$:

$$
g(5) = 9 - 2 \times 5 = 9 - 10 = -1
$$

Thus, $$ g(g(2)) = g(5) = -1 $$.

### (c) Find $$ f(g(x)) $$ in its simplest form

We need to substitute $$ g(x) = 9 - 2x $$ into $$ f(x) = x^2 + 6x + 5 $$:

$$
f(g(x)) = f(9 - 2x) = (9 - 2x)^2 + 6(9 - 2x) + 5
$$

Now, let's expand this step by step:

1. Expand $$ (9 - 2x)^2 $$:
   $$
   (9 - 2x)^2 = 81 - 36x + 4x^2
   $$

2. Expand $$ 6(9 - 2x) $$:
   $$
   6(9 - 2x) = 54 - 12x
   $$

3. Combine all parts:
   $$
   f(g(x)) = (81 - 36x + 4x^2) + (54 - 12x) + 5
   $$
   $$
   = 4x^2 - 36x - 12x + 81 + 54 + 5
   $$
   $$
   = 4x^2 - 48x + 140
   $$

Thus, $$ f(g(x)) = 4x^2 - 48x + 140 $$.

### (d) Find $$ f^{-1} $$

To find the inverse of $$ f(x) = x^2 + 6x + 5 $$, we first rewrite it in terms of $$ y $$:

$$
y = x^2 + 6x + 5
$$

Next, we solve for $$ x $$:

1. Rearranging gives:
   $$
   x^2 + 6x + (5 - y) = 0
   $$

2. Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:
   - Here, $$ a = 1, b = 6, c = 5 - y $$.
   $$
   x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (5 - y)}}{2 \cdot 1}
   $$
   $$
   = \frac{-6 \pm \sqrt{36 - 20 + 4y}}{2}
   $$
   $$
   = \frac{-6 \pm \sqrt{16 + 4y}}{2}
   $$
   $$
   = -3 \pm \sqrt{4 + y}
   $$

Since $$ f(x) $$ is a parabola opening upwards, we take the positive root for the inverse:

$$
f^{-1}(y) = -3 + \sqrt{4 + y}
$$

### (e) Solve the equation $$ f(x) = 2x + 17 $$

We set up the equation:

$$
x^2 + 6x + 5 = 2x + 17
$$

Rearranging gives:

$$
x^2 + 6x - 2x + 5 - 17 = 0
$$
$$
x^2 + 4x - 12 = 0
$$

Now, we can solve this quadratic equation using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
Where $$ a = 1, b = 4, c = -12 $$:

$$
x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}
$$
$$
= \frac{-4 \pm \sqrt{16 + 48}}{2}
$$
$$
= \frac{-4 \pm \sqrt{64}}{2}
$$
$$
= \frac{-4 \pm 8}{2}
$$

Calculating the two possible values:

1. $$ x = \frac{4}{2} = 2 $$
2. $$ x = \frac{-12}{2} = -6 $$

Thus, the solutions are $$ x = 2 $$ and $$ x = -6 $$.

### Summary of Answers:
- (a) $$ g(2) = 5 $$
- (b) $$ g(g(2)) = -1 $$
- (c) $$ f(g(x)) = 4x^2 - 48x + 140 $$
- (d) $$ f^{-1}(y) = -3 + \sqrt{4 + y} $$
- (e) Solutions: $$ x = 2 $$ and $$ x = -6 $$
(a) $$ g(2) = 5 $$ (b) $$ g(g(2)) = -1 $$ (c) $$ f(g(x)) = 4x^2 - 48x + 140 $$ (d) $$ f^{-1}(y) = -3 + \sqrt{4 + y} $$ (e) $$ x = 2 $$ and $$ x = -6 $$

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