Shrink Or Stretch? The Effect Of F(1/3 X) On Graphs

When we talk about transforming functions, we're essentially discussing how changing the input or output of a function alters its graphical representation. Today, we're diving deep into a specific transformation: what happens to the graph of $f(x)$ when we change it to $f

$\\left(\\frac{1}{3} x\\right)$? This might sound a bit technical, but trust me, it's a fundamental concept in understanding how function graphs behave. We're going to explore this intriguing change, break down why it happens, and figure out if it results in a shrink or a stretch, and in which direction.

Understanding Function Transformations

Before we zoom in on our specific example, let's set the stage with a general understanding of function transformations. These are operations that move, resize, or reflect the graph of a parent function. Common transformations include shifting the graph up, down, left, or right, stretching or compressing it vertically or horizontally, and reflecting it across an axis. Each of these transformations corresponds to a specific change in the function's equation. For instance, $f(x) + c$ shifts the graph vertically, while $f(x - c)$ shifts it horizontally. Reflections involve negative signs, like $-f(x)$ for a vertical reflection and $f(-x)$ for a horizontal reflection. Stretching and compressing are where things get particularly interesting, and our specific transformation falls into this category.

**The Core Question: What Does $f

$\\left(\\frac{1}{3} x\\right)$ Do?**

Our focus is on the transformation $f(x) o f

$\\left(\\frac{1}{3} x\\right)$. To understand this, let's think about what the 'x' in the function represents. It's the input value. When we replace 'x' with '\frac{1}{3}x', we are changing the input that the function $f$ receives. Let's consider a specific point on the graph of $f(x)$, say $(x_0, y_0)$, where $y_0 = f(x_0)$. Now, we want to find the corresponding point on the graph of $g(x) = f

$\\left(\\frac{1}{3} x\\right)$. For the output of $g(x)$ to be the same as $y_0$, we need $f

$\\left(\\frac{1}{3} x\\right) = y_0$. Since $y_0 = f(x_0)$, this means we need $f

$\\left(\\frac{1}{3} x\\right) = f(x_0)$. For this equality to hold, the input to the function $f$ must be the same. Therefore, we require $\\frac{1}{3}x = x_0$. Solving for $x$, we get $x = 3x_0$. This tells us that to get the same output $y_0$, the new input value $x$ must be three times the original input value $x_0$. If the input values are getting larger to produce the same output, this implies that the graph is being stretched horizontally. Imagine a point at $x=1$ on $f(x)$. To get the same height on $f

$\\left(\\frac{1}{3} x\\right)$, we need to input $x=3$, because $\\frac{1}{3}(3) = 1$. The point $(1, f(1))$ on the original graph is now at $(3, f(1))$ on the new graph. The x-coordinate has tripled, indicating a horizontal stretch.

Horizontal Shrink vs. Horizontal Stretch

It's crucial to distinguish between horizontal shrinks and stretches. A horizontal shrink occurs when the input values are made smaller to achieve the same output. This typically happens with transformations like $f(kx)$ where $k > 1$. For example, $f(3x)$ would shrink the graph horizontally because you need a smaller $x$ to get the same output (e.g., if $f(1)$ was a point, for $f(3x)$ to equal $f(1)$, we need $3x=1$, so $x=1/3$). Conversely, a horizontal stretch occurs when the input values are made larger to achieve the same output. This is precisely what we observe with $f

$\\left(\\frac{1}{3} x\\right)$. Since the coefficient of $x$ inside the function is $\\frac{1}{3}$ (which is less than 1), we are effectively dividing the input by $\\frac{1}{3}$, which is the same as multiplying it by 3. This multiplication of the input leads to a horizontal stretch.

Vertical Shrink vs. Vertical Stretch

Let's also clarify why this transformation does not affect the graph vertically. Vertical transformations involve changes to the output of the function, typically in the form of c ullet f(x) or $f(x) + c$. In $f

$\\left(\\frac{1}{3} x\\right)$, the operation is applied inside the function, directly modifying the input variable 'x'. The output $f

$\\left(\\frac{1}{3} x\\right)$ is still directly related to the output of the original function $f$. If $f(x)$ produces a value $y$, then $f

$\\left(\\frac{1}{3} x\\right)$ will also produce a value that is directly computed by $f$. There's no external multiplication or addition to the $f(x)$ term that would scale the y-values up or down. For example, if we had $2f(x)$, this would stretch the graph vertically by a factor of 2, because every y-value is doubled. Similarly, $\\frac{1}{2}f(x)$ would shrink the graph vertically. Since our transformation $f

$\\left(\\frac{1}{3} x\\right)$ doesn't have any multiplier outside the $f(\\cdot)$ or any additive constant, the vertical dimension of the graph remains unchanged.

Illustrative Example: $f(x) = x^2$

To make this concrete, let's consider the parent function $f(x) = x^2$. Its graph is a parabola opening upwards, with its vertex at the origin $(0,0)$. Now, let's look at the transformed function $g(x) = f

$\\left(\\frac{1}{3} x\\right) = \\left(\\frac{1}{3} x\\right)^2 = \\frac{1}{9} x^2$. Let's compare some points:

• For $f(x) = x^2$: At $x=3$, $f(3) = 3^2 = 9$. The point is $(3, 9)$.
• For $g(x) = \\frac{1}{9} x^2$: To get the same output of 9, we need $\\frac{1}{9} x^2 = 9$, which means $x^2 = 81$, so $x =

$\pm 9$. The points are $(9, 9)$ and $(-9, 9)$.

Notice how the x-coordinates have increased from $

$\pm 3$ to $

$\pm 9$. This is a multiplication by 3, confirming a horizontal stretch by a factor of 3. The y-coordinates (9 in this case) remain the same for these corresponding points. If we consider the vertex at $(0,0)$: For $f(x)$, $f(0)=0$. For $g(x)$, $g(0) = \\frac{1}{9}(0)^2 = 0$. The vertex remains at $(0,0)$, which is expected for horizontal transformations that don't involve shifts.

Connecting to the Options

Based on our analysis, the transformation $f(x) o f

$\\left(\\frac{1}{3} x\\right)$ causes the input values to be multiplied by 3 to achieve the same output. This means the graph is spread out along the x-axis. Therefore, the correct answer is that it shrinks it horizontally. Wait, did I get that right? Let's re-read our logic. We found that the new x-value is 3 times the old x-value to get the same y-value. This means the graph is wider. A wider graph is a stretch, not a shrink. My apologies, it's easy to get confused with these! So, to achieve the same output $y_0$, the new input $x$ needs to be $3x_0$. This means the graph is stretched horizontally. The input $\\frac{1}{3}x$ is 'diluted' compared to $x$, so to get the same effect from $f$, the input must be larger. This larger input stretches the graph horizontally.

Let's re-examine the options with this clarity:

A. shrinks it horizontally - This would mean the graph gets narrower, which requires smaller x-values for the same y-output. Our transformation leads to larger x-values.

B. shrinks it vertically - This would affect the y-values, making them smaller. Our transformation only affects the input 'x'.

C. stretches it vertically - This would affect the y-values, making them larger. Our transformation only affects the input 'x'.

D. stretches it horizontally - This means the graph gets wider, which requires larger x-values for the same y-output. This matches our finding that $x = 3x_0$ is needed to get the same $y_0 = f(x_0)$ when considering $f

$\\left(\\frac{1}{3} x\\right)$.

Therefore, the transformation $f(x) o f

$\\left(\\frac{1}{3} x\\right)$ stretches it horizontally. The factor of the horizontal stretch is the reciprocal of the coefficient of $x$ inside the function. In this case, the coefficient is $\\frac{1}{3}$, so the stretch factor is $\\frac{1}{\\frac{1}{3}} = 3$.

Key Takeaway: The Inside Game

Remember that transformations happening inside the function, directly modifying the input variable 'x' (like $f(kx)$ or $f(x+c)$), affect the graph horizontally. Transformations happening outside the function, applied to the entire function output (like k ullet f(x) or $f(x) + c$), affect the graph vertically. In our case, $f

$\\left(\\frac{1}{3} x\\right)$, the $\\frac{1}{3}$ is inside, so it's a horizontal change. Since the coefficient $\\frac{1}{3}$ is between 0 and 1, it leads to a stretch. If the coefficient were greater than 1 (e.g., $f(3x)$), it would lead to a shrink.

Conclusion

In summary, when you see a transformation of the form $f(x) o f(kx)$, you should immediately recognize it as a horizontal transformation. If $|k| > 1$, it's a horizontal shrink by a factor of $\\frac{1}{|k|}$. If $0 < |k| < 1$, it's a horizontal stretch by a factor of $\\frac{1}{|k|}$. For our specific question, $f(x) o f

$\\left(\\frac{1}{3} x\\right)$, we have $k = \\frac{1}{3}$. Since $0 <

$\\frac{1}{3} < 1$, this results in a horizontal stretch by a factor of $\\frac{1}{\\frac{1}{3}} = 3$. The graph becomes three times wider.

For further exploration into function transformations and graphical analysis, you might find the resources at Khan Academy helpful. They offer comprehensive guides and practice exercises on this topic.