To describe how to use function transformations to get from $\ln(x)$ to $f(x) = \ln(3-x) + 2$, we can break down the process into several steps, each corresponding to a standard transformation of the function $\ln(x)$.

Step-by-Step Function Transformations

1. Start with $\ln(x)$: This is the basic natural logarithm function with a domain of $(0, \infty)$.

2. Horizontal Reflection and Translation ( $\ln(-x)$ ):

   • Reflect $\ln(x)$ across the $y$-axis to get $\ln(-x)$.
   • The domain changes to $(-\infty, 0)$, which includes all negative values of $x$.

3. Horizontal Shift ( $\ln(3-x)$ ):

   • Shift $\ln(-x)$ to the right by 3 units. This gives the function $\ln(3-x)$.
   • The domain now changes to $(-\infty, 3)$, so the function is defined for all $x$ less than 3.

4. Vertical Translation ( $\ln(3-x) + 2$ ):

   • Shift $\ln(3-x)$ upward by 2 units. This results in the function $f(x) = \ln(3-x) + 2$.

Transformations Summary

• Reflection: $\ln(x)$ → $\ln(-x)$ (across the $y$-axis).
• Horizontal Shift: $\ln(-x)$ → $\ln(3-x)$ (right by 3 units).
• Vertical Shift: $\ln(3-x)$ → $\ln(3-x) + 2$ (up by 2 units).

Sketching the Graph of $f(x) = \ln(3-x) + 2$

1. Graph of $\ln(x)$:

   • The function $\ln(x)$ has a vertical asymptote at $x = 0$ and increases without bound as $x$ becomes larger.
   • It passes through $(1, 0)$ and approaches $-\infty$ as $x$ approaches 0 from the right.

2. Transform to $\ln(-x)$:

   • Reflect the graph of $\ln(x)$ across the $y$-axis.
   • Now there is a vertical asymptote at $x = 0$, but the function is defined for $x < 0$.

3. Transform to $\ln(3-x)$:

   • Shift the graph of $\ln(-x)$ 3 units to the right.
   • The vertical asymptote moves to $x = 3$.
   • The function passes through $(2, 0)$ and is defined for $x < 3$.

4. Transform to $\ln(3-x) + 2$:

   • Shift the graph of $\ln(3-x)$ upwards by 2 units.
   • Now the vertical asymptote remains at $x = 3$, and the graph passes through $(2, 2)$.
   • The graph approaches 2 as $x$ becomes very small (approaching $-\infty$).

Final Graph of $f(x) = \ln(3-x) + 2$

• Domain: $(-\infty, 3)$
• Vertical Asymptote: $x = 3$
• Horizontal Shift: Moves right by 3 units.
• Vertical Shift: Moves up by 2 units.

Would you like a detailed sketch of the graph or have any further questions on this topic?

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Here are 5 related questions for further exploration:

1. What is the range of the function $f(x) = \ln(3-x) + 2$?
2. How would you find the x-intercept of $f(x)$?
3. What is the behavior of $f(x)$ as $x$ approaches 3 from the left?
4. Can you find the inverse of the function $f(x)$?
5. How would you apply transformations if the function were $f(x) = -\ln(3-x) + 2$?

Tip: Understanding transformations of the basic logarithmic function is crucial for sketching more complex log functions effectively.