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Composition of Functions in Math-interactive

What kind of notation ?

Notation Meaning
$$ g \cdot f (x)$$ "g of f of x"
$$ g ( f (x) ) $$ "g of f of x"

Step By Step Flow Chart

Are compositions commutative?

Does $$ f(g(x)) = g(f (x)) $$?

Remember that the commutative property states that "order does not matter". ( link )

Multiplication is commutative because $$ 3 \cdot 2 = 2 \cdot 3 $$.

On the other hand, subtraction is not commutative because $$ 3 - 2 \ne 2 - 3 $$.

Does this property apply to compositions ?

In other words, does it make if difference whether we write $$ f(g(x)) $$ vs $$ g(f(x))$$ ?

To answer that question, let's explore a specific example:

Let

$$ \red{ f(x) = x + 5} $$

$$ \blue{ g(x) = 4x} $$

Let's evaluate $$ \red {f}(\blue{g}(3)) $$ and $$ \blue{g}(\red {f}(3)) $$ and see if we get the same answer.


$$ \red {f}(\blue{g}(3)) $$

$$ \red {f}(\blue{g}(3)) \\ \blue{g}(3) =4 \cdot 3 = 12 \\ \red {f}(12) = 12 + 5 \\ \red {f}(12) = \boxed {17 } $$

$$ \blue{g}(\red {f}(3)) $$

$$ \blue{g}(\red {f}(3)) \\ \red {f}(3) = 3+ 5 = 8 \\ \blue{g}(8) = 4 \cdot 8 \\ \blue{g}(8) = \boxed {32 } $$


Conclusion

Composition of functions are not commutative because $$ f(g(3)) \red { \ne} g(f(3)) $$.

What about the composition of inverse function

In other words, what if $$f(x)$$ and $$ g(x)$$ are inverses?

Again, the best way to understand this is to try some examples, and see what happens.

Example 1

Let $$ \bf{ \red { f(x) = x + 3}} \\ \bf{ \blue{ g(x) = f^{-1}(x) = x -3}} $$

What is $$ f(g(x)) = ? $$

$$ \red {f(}\blue{g(x)} \red {)} \\ = \red {f(}\blue{ x-3} \red {)} \\ = \red {(}\blue{ x-3} \red {) +3}= \bf{ x} $$

Example 2

Let $$ \bf{ \red { f(x) = x^2}} \\ \bf{ \blue{ g(x) = f^{-1}(x) = \sqrt{x}}} $$

What is $$ f(g(x)) = ? $$

$$ \red {f(}\blue{g(x)} \red {)} \\ =\red {f(}\blue{ \sqrt{x}} \red {)} \\ = \red {(} \blue{ \sqrt{x}} \red {)^2} = \bf{ x} $$


Conclusion

Composition of inverse functions always evaluate to $$ x $$ (ie the input itself) .

$$ f(f^{-1}(x)) = \bf {x} $$

Therefore, you know that:

$$ g(g^{-1}(5)) = 5 \\ a(a^{-1}(232232)) = 232232 \\ k(k^{-1}(-90210)) = -90210 \\ k(k ^{-1}(\bigstar)) = \bigstar \\ f(f^{-1}(stuff)) = stuff $$

If you think of the inverse function as 'undoing' the original function, this rule might make sense. Read more about inverse functions here.

Practice Problems Part I

Directions: Evaluate each composition of functions listed below.

Problem 1

Let $$ f(x) = x + 3$$ and $$ g(x) = 4x $$. Evaluate $$ f(g(2)) $$.

Step 1

Evaluate function $$ f(\red {g(2) }) $$.

$$ g(2) = $$
$$ 4(2) = \blue{8} $$
Step 2

Substitute that value into left side function $$ \red {f}(g(2)) $$.

$$ f(\blue{ 8 }) $$
Step 3
$$ f(\blue{ 8 }) $$ =
$$ 8 + 3 = 11 $$
Problem 2

Let $$ f(x) = x^2 $$ and $$ g(x) = x+1 $$. Evaluate $$ (f \cdot g)(8) $$.

Remember: $$(f \cdot g)(8)$$ is the same as $$ f(g(8)) $$.

Step 1

Evaluate function $$ f(\red {g(8) }) $$.

$$ g(8) = $$
$$ 8+1 = \blue{9} $$
Step 2

Substitute that value into left side function $$ \red {f}(g(8)) $$.

$$ f(\blue{ 9 }) $$
Step 3
$$ f(\blue{ 9 }) $$ =
$$ 9^2 = 81 $$
Problem 3

Let $$ f(x) = | 7x + 4 | $$ and $$ h(x) = \sqrt{2x}$$. Evaluate $$ (f \cdot h)(18) $$.

Remember: $$(f \cdot h)(18)$$ is the same as $$ f(h(18)) $$.

Step 1

Evaluate function $$ f(\red {h(18) }) $$.

$$ h(18) = $$
$$ \sqrt{2\cdot 18} =\sqrt{36}= \blue{6} $$
Step 2

Substitute that value into left side function $$ \red {f}(h(18)) $$.

$$ f(\blue{ 6 }) $$
Step 3
$$ f(\blue{ 6 }) $$ =
$$ | 7\cdot 6 + 4 | = 46 $$
Problem 4

Let $$ f(x) = 3x -1 $$ and $$ b(x) = x^3 $$. Evaluate $$ f(b(5)) $$.

Step 1

Evaluate function $$ f(\red {b(5) }) $$.

$$ b(5) = $$
$$ 5 ^3= \blue{125} $$
Step 2

Substitute that value into left side function $$ \red {f}(b(5)) $$.

$$ f(\blue{ 125 }) $$
Step 3
$$ f(\blue{ 125 }) $$ =
$$ 3(\blue{125})-1 = 374 $$
Problem 5

Let $$ f(x) = \frac{1}{3}x + 2 $$ and $$m(x) = 3x - 6$$. Evaluate $$ f(m(11)) $$.

Step 1

Evaluate function $$ f(\red {m(11) }) $$.

$$m(11) = $$
$$ 5 ^3= \blue{125} $$
Step 2

Substitute that value into left side function $$ \red {f}(m(11)) $$.

$$ f(\blue{ 27 }) $$
Step 3
$$ f(\blue{ 27 }) $$ =
$$ \frac{1}{3} \blue{27} + 2 = 9 + 2=11 $$

Did you notice that the input of '11' is the same as the output for this composition.

Why do you think that is?

Because these two functions are inverses of each other!
Remember that the rule about the composition of inverse functions.

Exclamation mark

Advanced problems - solving algebraically.

Practice Problems Part II

Problem 1

Let $$ k(x) = x^2 + x $$ and $$ h(x) = x + 3$$. Evaluate $$ (k \cdot h)(x) $$.

Remember: $$(\red {k} \cdot \blue{h})(x)$$ is the same as $$ \red {k} (\blue{h} (x)) $$.

Step 1

Substitute right side function into left side.

$$ \red { k(} \blue{ h(x) } \red {)} = \red { k(} \blue{ x + 3 } \red {)} \\ \red { k(} \blue{ x + 3 } \red {)} = \red {(} \blue{ x + 3 } \red {)^2 + (}\blue{ x + 3 } \red {)} $$

Step 2

Simplify.

$$ \red {(}x^2 + 6x + 9 \red {) + (} x + 3 \red {)} \\ x^2 + 6x + 9 + x + 3 \\ x^2 + 7x + 12 $$

Problem 2

Let $$ g(x) = 2x^2 - 3x$$ and $$ h(x) = x + 4$$. Evaluate $$ (g \cdot h)(x) $$.

Remember: $$(\red {g} \cdot \blue{h})(x)$$ is the same as $$ \red {g} (\blue{h} (x)) $$.

Step 1

Substitute right side function into left side.

$$ \red { g(} \blue{ h(x) } \red {)} = \red { g(} \blue{ x + 4 } \red {)} \\ \red { g(} \blue{ x + 4 } \red {)} = \red {2(} \blue{ x + 4 } \red {)^2 - 3(}\blue{ x + 4 } \red {)} $$

Step 2

Simplify.

$$ \red {2(} \blue{ x + 4 } \red {)^2 - 3(}\blue{ x + 4 } \red {)} \\ 2 (x^2 + 8x + 16) - 3 \cdot x - 3 \cdot 4 \\ 2x^2 + 2\cdot 8x + 2\cdot 16 - 3 x - 12 \\ 2x^2 + 16x + 32 - 3x - 12 \\ 2x^2 + 13x +20 $$


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