Section 2.5: Pre-Composition and Graphing

In a previous section, we examined how applying arithmetic operations after a given function affects it algebraically, numerically, and geometrically.  We illustrated the resulting graphing principles by examining linear functions.  Now we want to investigate the effects of applying operations before a given function, that is, by pre-composition.

As before, we will investigate the effects of pre-composition by using the function, k, going through the points:

Now consider how pre-composition with a function, f, affects this function numerically and algebraically.  Algebraically, pre-composition gives , that is, the function f is applied before (on the "inside" of) k.  Numerically, computing simply involves applying k to the outputs of f.  This makes it a bit more difficult to compute a table of values for .  We must compute those inputs of f which will evaluate to the given input values for k; that is, we must apply the inverse, f ^-1, to the inputs of k.

For example, consider pre-composition with each of the arithmetic operations from before:

• Addition: if f(x) = x + 1, then , and we should subtract 1 (since f ^-1(x) = x - 1) from the inputs of k:

  x	y = k(x + 1)
  -5 -2 -1 0 3	1 -1 0 1 -3

  Check: For example, plugging in x = -5, gives k(-5 + 1) = k(-4) = 1.  Notice that, since we are replacing the input x in the original formula by x + 1, it makes sense that we should change the input values.
• Multiplication by a positive number: if g(x) = 2x, then , and we should divide the inputs of k by 2 (since g ^-1(x) = x/2):

  x	y = k(2x)
  -2 -1/2 0 1/2 2	1 -1 0 1 -3

  Check: For example, plugging in x = -2, gives k(2(-2)) = k(-4) = 1.
• Negation: if h(x) = -x, then , and we should negate the inputs of k (since h ^-1(x) = h(x) = -x):

  x	y = -k(x)
  4 1 0 -1 -4	1 -1 0 1 -3

  Check: For example, plugging in x = 4, gives k(-(4)) = k(-4) = 1.

Since only the input values in each case have been affected, which are plotted horizontally, each of these pre-compositions should affect the graph horizontally, by transforming the horizontal axis.  However, since the function is affected numerically by the inverse operations, we should expect the graph to be affected geometrically by the corresponding inverse operations.

For example, since the opposite of "add 1" is "subtract 1"", which shifts the number line by 1 unit in the negative direction, the graph of y = k(x + 1) should look like the graph of y = k(x), but shifted left one unit.  We can use XFunctions to quickly verify this.

1. Use the following applet:

   and select "Example: k(x) vs. k(x+1)" from the menu at the top of the applet.

as predicted, and the second graph is shifted left by 1 unit.

Similarly, the opposite of "multiply by 2" is "divide by 2", which shrinks the number line by a factor of 2.  This suggests that the graph of y = k(2x) should look like the graph of y = k(x), but shrunk horizontally by a factor of 2.  

2. Use the following applet for the next experiment. 

3. Select "Example: k(x) vs. k(2x)" from the menu at the top of the applet.

as predicted, and the second graph is shrunk horizontally (towards the the y-axis) by a factor of 2.  Notice: Although k(2x) looks algebraically very similar to 2k(x), the effect of the factor of 2 is exactly opposite (horizontal vs. vertical, shrink vs. stretch).  In general:

We can also verify that negation acts as expected.  Since the opposite of "negation" is "negation", which flips the number line, we expect the graph of y = k(-x) to look like the graph of y = k(x), but flipped horizontally.  

4. Use the following applet for the next experiment.  

5. Select "Example: k(x) vs. k(-x)" from the menu at the top of the applet.

as predicted, and the second graph is flipped horizontally (around the y-axis).

Since pre-composition seems to work "backwards", it might not be such a surprise to see what happens when we apply a series of operations.  For example, consider r(x) = -2x + 1, (i.e., : " take the input, multiply by 2, negate this, and add 1").  If we try to relate k(x) to k(-2x + 1) one step at a time, we can see that we should pre-compose with g, h, and f in reverse order; that is, start with k(x), then:

• pre-compose with f: , which corresponds to subtracting 1 from the inputs/shifting the graph left by 1, then
• pre-compose with h: , which which corresponds to negating the inputs/flipping the graph horizontally, then
• pre-compose with g: , which corresponds to dividing the inputs by 2/shrinking horizontally by a factor of 2.

6. You can see this using the following applet:  

7. This will compare each successive pair of graphs; simply select each successive example from the top menu.  

This behavior is actually not so strange, since we have already seen that, to invert a series of operations, we must apply the opposite of each operation in the reverse order.  For example, since r(x) = -2x + 1 takes the input:

• multiplies by 2, 
• negates this, and 
• adds 1, 

to apply r ^-1 to the inputs of k, we would need to:

• subtract 1 from the input, then
• negate this and
• divide by 2.

This gives:

so that 

8. You can use the above applet  to verify that these are points on the graph of k(-2x + 1) and that they correspond to the original points of k(x); simply select the Example from the top menu.

To summarize, we have seen that:

Notice that the geometric and numerical effects directly correspond, in that they both correspond to the reverse of the algebraic operation applied to the input variable, x.  As with post-composition, while this principle applies to any functional operation, we will use this graphing principle primarily in these three cases of multiplication and addition with constants.

On the following worksheet, we will refine our earlier technique to analyze a known function composed with arithmetic operations to create an accurate sketch.

Go to Transformational Graphing

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