Combining Functions

Consider the functions \(f(x) = 2x + 1\) and \(g(x) = 3x +2\).

What happens when we combine them by addition?

\[(f+g)(x) = f(x) + g(x)\]

Now we can substitute each of the functions.

\[(f + g)(x) = (2x +1) + (3x +2)\]

And finally, collect like terms.

\[(f + g)(x) = 5x + 3\]

Let's consider the same functions again, \(f(x) = 2x + 1\) and \(g(x) = 3x +2\).

What happens when we combine them by subtraction?

\[(f-g)(x) = f(x) - g(x)\]

Now we substitute the functions as before.

\[(f - g)(x) = (2x +1) - (3x +2)\]

And we can collect like terms.

\[(f - g)(x) = -x - 1\]

Let's consider the functions \(f(x) = 2x + 4\) and \(g(x) = x + 2\)

What happens if we multiply them together.

\[(fg)(x) = f(x) \cdot g(x)\]

And substitute in the functions.

\[(fg)(x) = (2x + 4)(x+2)\]

We then multiply out the brackets.

\[(fg)(x) = 2x^2 + 4x + 4x + 8\]

And again we can collect like terms.

\[(fg)(x) = 2x^2 + 8x + 8\]

Let's consider the functions \(f(x) = 2x + 4\) and \(g(x) = x + 2\) once again.

What happens when we combine them by division.

\[(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\]

And substitute in the functions.

\[(\frac{f}{g})(x) = \frac{2x+4}{x+2}\]

Then we simplify algebraically. In the case of these functions, we find a common factor and divide through.

\[(\frac{f}{g})(x) = \frac{2(x+2)}{x+2}\]

\[(\frac{f}{g})(x) = 2\]

Let's say we have the functions:

\[f(x) = \sqrt x g(x) = \sqrt{2-x}\]

The domain of f(x) is \(A = [0, \infty )\).

The domain of g(x) is \(B = (-\infty, 2]\).

So, the domain of (f + g)(x) is:

\((f + g) (x) = \sqrt x + \sqrt{2-x}, \text{ Domain of } (f + g)(x) \text{ is }A \cap B = [0, 2]\)

Similarly, the domain of (f - g)(x) is:

\((f - g) (x) = \sqrt x - \sqrt{2-x}, \text{ Domain of } (f - g)(x) \text{ is }A \cap B = [0, 2]\)

Let's say we have the functions:

\[f(x) = x^2; \space g(x) = x-1\]

The domain of f(x) is \(A = (-\infty, \infty)\).

The domain of g(x) is \(B = (-\infty, \infty)\).

So, the domain of \((f \cdot g) (x)\) is:

\((fg)(x) = x^2(x-1); \text{ Domain of } (fg)(x) \text{ is } A\cap B = (-\infty, \infty)\)

But, the domain of \((\frac{f}{g})(x)\) is:

\((\frac{f}{g})(x) = \frac{x^2}{x-1}; \text{ Domain of } (\frac{f}{g})(x) \text{ is } A \cap B= (-\infty, 1)\cup (1, \infty)\)

\[f(4) = 5(4) + 2 = 20 + 2 = 22\]

\[g(4) = (4)^2 - 1 = 16-1 = 15\]

Let's say we have two functions:

\(f(x) = 2 + 3x - x^2\) and \(g(x) = 2x -1\)

What is \((f \circ g)(x)\)?

Solution:

1. The first step is to understand what \((f \circ g)(x)\) is telling us to do. Written in this order, we need to plug the function g(x) into the function f(x) and solve.
   • From there, we need to remember the formula and write it down. So,
   • \((f \circ g)(x) = f(g(x))\)
2. Since we know the formula for g(x), let's plug it in:
   • \(f(g(x)) = f(2x-1)\)
3. Now, everywhere in the function f, replace x with 2x - 1:
   • \((f \circ g)(x) = f(2x-1) = 2+3(2x-1)-(2x-1)^2\)
4. And simplify:
   • \((f \circ g)(x) = 2 + 3(2x-1) - (2x-1)^2 \rightarrow (f \circ g)(x) = 2 + 6x-3-(4x^2-4x+1)\rightarrow (f\circ g)(x) = -1 + 6x-4x^2 + 4x-1 \rightarrow (f \circ g)(x) = -4x^2 + 10x -2\)

Let's take the same two functions in the previous example: \(f(x) = 2 + 3x - x^2\) and \(g(x) = 2x -1\)

Only this time, let's find \((g \circ f)(x)\).

Solution:

1. Again, we first need to understand what \((g \circ f)(x)\) is telling us to do. Written in this order, we need to plug the function f(x) into the function g(x) and solve.
   • So let's write the formula:
   • \((g \circ f)(x) = g(f(x))\)
2. Since we know the formula for f(x), let's plug it in:
   • \(g(f(x)) = g(2 + 3x - x^2)\)
3. Now, everywhere in the function g, replace x with \(2+3x-x^2\):
   • \((g \circ f)(x) = g(2+3x-x^2) = 2(2 + 3x - x^2) - 1\)
4. And simplify:
   • \((g \circ f)(x) = 2(2 + 3x - x^2) - 1 \rightarrow (g \circ f)(x) = 4 + 6x - 2x^2 -1 \rightarrow (g \circ f)(x) = -2x^2 + 6x + 3\)

This looks like: \((f \circ f)(x) = f(f(x))\)

So, if we have a function:

\[f(x) = 2x +3\]

and compose it with itself, we get:

\((f \circ f)(x) = f(f(x)) = f(2x + 3) = 2(2x+3)+3 = 4x + 6 +3 = 4x + 9\)

The domain of \(f(x) = \sqrt{x}\) is \([0, \text{ is } \infty)\).

The domain of \(g(x) = x^2\) is \((-\infty, \infty)\)

If we compose them as:

\((f \circ g)(x) = f(g(x))\)

we get:

\(f(g(x)) = f(x^2) = \sqrt{x^2} = x\)

• In this case, since both x and g(x) have a domain of all real numbers:
  • The domain is all real numbers: \((-\infty, \infty)\)

But, if we compose them as:

\[(g \circ f)(x) = g(f(x))\]

we get:

\[g(f(x)) = g(\sqrt x) = (\sqrt{x})^2 = x\]

• In this case, while x would normally have a domain of all real numbers, we must consider f(x), which has a domain of all non-negative real numbers, so:
  • The domain is all non-negative real numbers: \([0, \infty)\)

Let's write the function:

\[f(x) = \sqrt{5 -x^2}\]

as the composition of two simpler functions.

Solution 1:

We are trying to break up, or decompose, f(x) into two simpler functions that we will call g(x) and h(x). So,

\[f(x) = g(h(x))\]

1. To break up the function, we first must look for a function inside a function within f(x).
   • As one option, we might see that the inside function is \(5-x^2\), and the outside function is \(\sqrt x\).
2. After observing this, we could decompose the function as
   • \(h(x) = 5 -x^2; \space g(x) = \sqrt x\)
3. The final step is to double-check our answer by recomposing the functions:
   • \(g(h(x)) = g(5 -x^2) = \sqrt{5-x^2} = f(x)\)

Solution 2:

1. Another option would be to break up f(x) into g(x) and h(x) as:
   • \(h(x) = x^2g(x) = \sqrt{5-x}\)
2. Again, we need to double-check our answer by recomposing the functions:
   • \(g(h(x)) = g(x^2) = \sqrt{5-x^2} = f(x)\)

Say we want to compose two functions:

\[f(x) = \sqrt x; \space g(x) = x-1\]

If we want to compose them as \((f \circ g)(x)\):

1. Write the formula - \((f \circ g)(x) = f(g(x))\)
   • This makes it more clear that
     • f(x) is the "outer" function and
     • g(x) is the "inner" function
2. Now, replace the variable, x, of the "outer" function, f(x), with the "inner" function, g(x) to get:
   • \((f \circ g)(x) = f(g(x)) = f(x-1) = \sqrt{x-1}\)
   • So, the composed function is: \(f(g(x)) = \sqrt{x-1}\)

If we compare the composition, \((f \circ g)(x)\), to the multiplication, \((f \cdot g)(x)\), (this could also be written as (fg)(x) we see that the answers are not the same.

• When we multiply \((f \cdot g)(x)\), we get:
  • \((f \cdot g)(x) = \sqrt x \cdot (x-1) = (\sqrt x)(x) - \sqrt x = x \sqrt x - \sqrt x\)
  • Which is not the same as the composition. So, \((f \circ g) (x) ≠ (f \cdot g)(x)\)!

Given f, g, and h: \((f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(g(h(x))) = (f \circ g)(h(x)) = ((f \circ g) \circ h)(x)\)

So, \(f \circ(g \circ h) = (f \circ g) \circ h\)

Say we have two piecewise functions:

\(f(x) = \begin{cases} 2 & x > 2 \\ 2 & x < 2 \end{cases}\) and \(g(x) = \begin{cases} -2 & x \geq 2 \\ 2 & x<2 \end{cases}\)

And we want to calculate g(x) - f(x).

Solution:

For this piecewise function, we need to split it into cases.

• For \(x <2\):
  • f(x) = 3 and g(x) = 2.
    • So, \((g - f)(x) = -1\)
• At \(x = 2\):
  • f(x) doesn't exist
    • So, \((g - f)(x)\) is undefined
• For \(x > 2\):
  • \(f(x) = 2\) and \(g(x) = -2\).
    • So, \((g - f)(x) = -4\)

If \(f(x) = x^2 - 2x - 3\) and \(g(x) = x + 1\), find the following:

1. \((f-g)(-1)\)
2. \(\left( \frac{f}{g} \right) (x)\)

Solutions:

1. Here we need to subtract g(x) from f(x) and then substitute -1 for x.
   • \((f-g)(x) = f(x) - g(x) = (x^2 -2x -3)-(x+1) = x^2 -2x - 3 -x -1 = x^2 - 3x -4\)
   • \((f -g)(-1) = (-1)^2 -3(-1) -4 = 1 + 3 -4 = 0\)
2. Here we need to divide f(x) by g(x). Note that we can factor f, so the fraction can be simplified.
   • \(\left( \frac{f}{g} \right) = \frac {x^2-2x-3}{x+1} = \frac {(x-3)\cancel{(x+1)}}{\cancel{x+1}} = x-3 \text{ if } x≠ 1\)

Find \((f+g), (f-g), (fg),\text{ and } \left( \frac{f}{g} \right)\) for the functions:

\[f(x) = x^2 + 3g(x) = x-1\]

Solutions:

• \((f + g) = f(x) + g(x) = (x^2 + 3) + (x-1) = x^2 + 3 +x -1 = x^2 + x +2\)
• \((f-g) = f(x) - g(x) = (x^2+3) - (x-1) = x^2 + 3 - x +1 = x^2 - x + 4\)
• \((fg) = f(x) \cdot g(x) = (x^2 + 3)(x-1) = x^3 - x^2+3x-3\)
• \(\left( \frac{f}{g} \right) = \frac{f(x)}{g(x)} = \frac{x^2+3}{x-1}, x ≠ 1\)

The graph of two piecewise functions, StudySmarter Originals

The graph of the function f is shown in blue, and the graph of the function g is shown in green. Use the graphs to solve each quantity.

1. \(f(3)\)
2. \(g(3)\)
3. \(f(3) + g(3)\)
4. \((f -g) (3)\)

Solutions:

1. \(f(3) = -3\)
2. \(g(3) = 9\)
3. \(f(3) + g(3) = -3 + 9 = 6\)
4. \((f -g) (3) = f(3) - g(3) = -3 -9 = -12\)

If \(f(x) = x^2 + 10\) and \(g(x) = \sqrt{x-1}\), find the following:

1. \(f(g(x))\)
2. \((g \circ f)(4)\)
3. \((f\circ f)(x)\)

Solutions:

1. Here we need to substitute g(x) for the x in f(x).
   • \(f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 +10\)
   • \(f(g(x)) = x-1+10\)
2. Here we need to find g(f(x)) and then evaluate at x = 4.
   • \((g \circ f)(x) = g(f(x)) = g(x^2 + 10) = \sqrt{(x^2 + 10) - 1} = \sqrt{x^2 + 9}\)
   • \((g \circ f)(4) = \sqrt{(4)^2 + 9} = \sqrt{16+9} = \sqrt{25} = 5\)
3. Here we are composing f with itself, so we substitute \(x^2-10\) for the x in f.
   • \((f \circ f)(x) = f(f(x)) = f(x^2 + 10) = (x^2 +10)^2 + 10\)
   • \((f \circ f)(x) = x^4 + 20 x^2 + 110\)

If \(f(x) = \sqrt x\) and \(g (x) = \sqrt{2-x}\), find each composition and its domain.

1. \((f \circ g)\)
2. \((g \circ f)\)
3. \((f \circ f)\)
4. \((g \circ g)\)

Solutions:

1. \(f \circ g = f(g(x))\)
   • \(f(g(x)) = f(\sqrt{2-x}) = \sqrt {\sqrt{2-x}} = \sqrt[4]{2-x}\)
   • The domain of \(f(g(x))\) is \((-\infty, 2]\)
2. \(g \circ f = g(f(x))\)
   • \(g(f(x))= g(\sqrt{x}) = \sqrt{2-\sqrt{x}}\)
   • To find the domain, both \(\sqrt x\) and \(\sqrt{2-\sqrt{x}}\) must have \(x \geq 0\).
     • For \(\sqrt{x}\) to be defined, \(x \geq 0\).
     • For \(\sqrt{2-\sqrt x}\) to be defined, \(2-\sqrt x \geq 0\), or \(\sqrt x \leq 2\), or \(x \leq 4\).
     • So, to meet both requirements, the domain of \(g(f(x))\) is \([0, 4]\)
3. \(f \circ f = f(f(x))\)
   • \(f(f(x)) = f(\sqrt{x}) = \sqrt {\sqrt{x}} = \sqrt[4]{x}\)
   • The domain of \(f(f(x))\) is \([0, \infty))
4. \(g \circ g = g(g(x))\)
   • \(g(g(x)) = g(\sqrt{2-x}) = \sqrt{2 - \sqrt{2-x}}\)
   • To find the domain, both \(2 - x \leq 0\) and \(2 - \sqrt{2-x} \geq 0\).
     • For \(2-x \geq 0, \space x \leq 2\)
     • For \(2 - \sqrt{2-x} \geq 0, \space \sqrt{2-x} \leq 2 \rightarrow 2 - x \leq 4 \rightarrow x \geq -2\)
     • So, the domain of \(g(g(x))\) is \([-2, 2]\)

We can also compose three or more functions! For instance, the composite function \(f \circ g \circ h\) can be found by first applying h, then g, and then f like so:

\[(f \circ g\circ h)(x) = f(g(h(x)))\]

So, find \((f \circ g \circ h)(x)\) if:

\[f(x) = \frac{x}{x+1} \qquad g(x) = x^{10} \qquad h(x) = x +3 \]

Solution:

\[(f \circ g \circ h)(x) = f(g(h(x))) = f(g(x+3)) = f \left( (x+3)^{10}\right) = \frac{(x+3)^{10}}{(x+3)^{10}+1}\]

Given the function:

\[F(x) = \cos^2(x+9)\]

Find functions f, g, and h such that \(F = f \circ g \circ h\).

Solution:

1. Since \(F(x) = (\cos(x+9))^2\), the formula for F says:
   1. First add 9
   2. Then take the cosine of the result
   3. And square it.
2. So, we let:
   1. \(h(x) = x +9\)
   2. \(g(x) = \cos(x)\)
   3. \(f(x) = x^2\)
3. Therefore, \((f \circ g \circ h)(x) = f(g(h(x))) = f(g(x+9)) = f(\cos(x+9)) = (\cos(x+9))^2 = F(x)\)