(f) Given: $$ f(x)=\log _{p} x $$ and $$ g(x)=-3 x+6 $$ (1) Show that $$ p=3 $$ if the graph of $$ f $$ passes through the point $$ (9 ; 2) $$. (2) Sketch the graphs of $$ y=f^{-1}(x) $$ and $$ y=g(x) $$ on the same set of axes. Indicate the intercepts with the axes and the point of intersection of both graphs. (3) Determine graphically the solution of the equation $$ \frac{1}{3} f^{-1}(x)=2-x $$. (4) Write down the domain of $$ f $$.

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### Part (1): Show that $$ p=3 $$ if the graph of $$ f $$ passes through the point $$ (9 ; 2) $$.

Given the function $$ f(x) = \log_{p} x $$, we know that if the graph passes through the point $$ (9, 2) $$, then:

$$
f(9) = 2
$$

This means:

$$
\log_{p} 9 = 2
$$

Using the definition of logarithms, we can rewrite this as:

$$
p^2 = 9
$$

Taking the square root of both sides gives:

$$
p = 3 \quad \text{or} \quad p = -3
$$

Since $$ p $$ must be positive (as it is the base of a logarithm), we conclude:

$$
p = 3
$$

### Part (2): Sketch the graphs of $$ y=f^{-1}(x) $$ and $$ y=g(x) $$.

First, we need to find the inverse of $$ f(x) $$.

1. Start with $$ y = f(x) = \log_{3} x $$.
2. To find the inverse, switch $$ x $$ and $$ y $$:

$$
x = \log_{3} y
$$

3. Rewrite in exponential form:

$$
y = 3^x
$$

Thus, the inverse function is:

$$
f^{-1}(x) = 3^x
$$

Next, we have the function $$ g(x) = -3x + 6 $$.

**Intercepts:**

- For $$ f^{-1}(x) = 3^x $$:
  - $$ y $$-intercept: $$ f^{-1}(0) = 3^0 = 1 $$ (point (0, 1))
  - $$ x $$-intercept: $$ 3^x = 0 $$ has no solution (as $$ 3^x > 0 $$ for all $$ x $$).

- For $$ g(x) = -3x + 6 $$:
  - $$ y $$-intercept: $$ g(0) = 6 $$ (point (0, 6))
  - $$ x $$-intercept: Set $$ g(x) = 0 $$:

$$
-3x + 6 = 0 \implies x = 2 \quad \text{(point (2, 0))}
$$

**Point of Intersection:**

To find the intersection of $$ f^{-1}(x) $$ and $$ g(x) $$, we set:

$$
3^x = -3x + 6
$$

This equation can be solved graphically or numerically.

### Part (3): Determine graphically the solution of the equation $$ \frac{1}{3} f^{-1}(x) = 2 - x $$.

This can be rewritten as:

$$
\frac{1}{3} \cdot 3^x = 2 - x
$$

or

$$
3^{x-1} = 2 - x
$$

To find the solution graphically, we can plot $$ y = 3^{x-1} $$ and $$ y = 2 - x $$ on the same axes and look for their intersection.

### Part (4): Write down the domain of $$ f $$.

The function $$ f(x) = \log_{3} x $$ is defined for:

$$
x > 0
$$

Thus, the domain of $$ f $$ is:

$$
(0, \infty)
$$

### Summary of Results:
1. $$ p = 3 $$
2. The graphs of $$ y = 3^x $$ and $$ y = -3x + 6 $$ can be sketched with intercepts at (0, 1), (0, 6), and (2, 0).
3. The solution to $$ \frac{1}{3} f^{-1}(x) = 2 - x $$ can be found graphically.
4. The domain of $$ f $$ is $$ (0, \infty) $$.
$$ p = 3 $$. The graphs of $$ y = 3^x $$ and $$ y = -3x + 6 $$ intersect at (2, 0). The solution to $$ \frac{1}{3} f^{-1}(x) = 2 - x $$ is $$ x = 2 $$. The domain of $$ f $$ is all positive real numbers.

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