The transformation $f(x) \mapsto \frac{1}{3}f(x)$ affects the vertical scaling of the graph of $f(x)$. Here's what happens:

• Vertical Compression: Every point on the graph of $f(x)$ has its $y$-coordinate multiplied by $\frac{1}{3}$. This reduces the height of the graph by a factor of $3$, effectively compressing it vertically towards the $x$-axis.
• Key Points: Points on the graph of $f(x)$ at $(x, y)$ will move to $(x, \frac{y}{3})$.

Summary:

The transformation $f(x) \mapsto \frac{1}{3}f(x)$ compresses the graph of $f(x)$ vertically by a factor of $3$. The shape and horizontal positioning remain unchanged.

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Additional Questions:

1. How would the transformation $f(x) \mapsto 3f(x)$ affect the graph?
2. What would happen if $f(x) \mapsto -\frac{1}{3}f(x)$?
3. How does a horizontal scaling transformation differ from vertical scaling?
4. Can you determine how critical points (like maxima and minima) change under this transformation?
5. How does this transformation affect the area under the curve if integrated over a fixed interval?

Tip:

To understand transformations visually, try sketching both $f(x)$ and the transformed version on the same coordinate plane.