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Composing functions (article), Lecture notes of Algebra

Composing functions Khan academy with examples and problems to practice

Typology: Lecture notes

2021/2022

Uploaded on 02/03/2022

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Walk through examples, explanaons, and pracce problems to
learn how to find and evaluate composite funcons.
Given two funcons, we can combine them in such
a way so that the outputs of one funcon become
the inputs of the other. This acon defines a
composite funcon. Let's take a look at what this
means!
Evaluang composite funcons
Example
If and , then what is
?
Soluon
One way to evaluate is to work from the
"inside out". In other words, let's evaluate first
and then substute that result into to find our
answer.
Let's evaluate .
Composing funcons
f(x) = 3x 1 g(x) = x+
32
f(g(3))
f(g(3))
g(3)
f
g(3)
pf3
pf4
pf5
pf8
pf9
pfa

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Walk through examples, explanaons, and pracce problems to learn how to find and evaluate composite funcons.

Given two funcons, we can combine them in such a way so that the outputs of one funcon become the inputs of the other. This acon defines a composite funcon. Let's take a look at what this means!

Evaluang composite funcons

Example

If and , then what is ?

Soluon

One way to evaluate is to work from the "inside out". In other words, let's evaluate first and then substute that result into to find our answer.

Let's evaluate.

Composing funcons

f (x) = 3x − 1 g(x) = x 3 + 2 f (g(3))

f (g(3)) g(3) f

g(3)

Since , then.

Now let's evaluate.

It follows that.

g(x)

g(3)

= x 3 + 2

= (3) 3 + 2 Plug in x =

g(3) = 29 f (g(3)) = f (29)

f (29)

f (x)

f (29)

= 3x − 1

= 3(29) − 1 Plug in x =

f (g(3)) = f (29) = 86

(^3) g 29 f 86

Since , we can substute in for .

This new funcon should take directly to. Let's verify this.

Excellent!

Let's pracce

Problem 1

g(x) = x 3 + 2 x 3 + 2 g(x)

f (g(x)) = 3(g(x)) − 1

= 3(x 3 + 2) − 1

= 3x 3 + 6 − 1

= 3x 3 + 5

f (g(x))

f (g(3))

= 3x 3 + 5

f (x) = 3x − 1

Problem 2

Composite funcons: a formal definion

In the above example, we found and evaluated a composite funcon.

g(x) = x 3 + 2

Evaluate g(f (1)).

Check

[I need help. Please show me the soluon.]

m(x) = 3x − 2

n(x) = x + 4

Find m(n(x)).

Check

[I need help. Please show me the soluon.]

We can find as follows:

Since we now have funcon , we can simply substute in for to find.

Of course, we could have also found by evaluang. This is shown below:

(h ∘ g)(x)

(h ∘ g)(x) = h(g(x))

= (g(x)) 2 − 2(g(x))

= (x + 4) 2 − 2(x + 4)

= x 2 + 8x + 16 − 2x − 8

= x 2 + 6x + 8

h ∘ g −2 x (h ∘ g)(−2)

(h ∘ g)(x)

(h ∘ g)(−2)

= x 2 + 6x + 8

(h ∘ g)(−2) h(g(−2))

The diagram below shows how is related to.

Here we can see that funcon takes to and then funcon takes to , while funcon takes directly to.

Now let's pracce some problems

Problem 3

(h ∘ g)(−2) = h(g(−2))

= h(2) Since g(−2) = −

= 0 Since h(2) = 2^2

(h ∘ g)(−2) h(g(−2))

−2 (^) g 2 h 0

h ∘ g

g −2 2 h 2 0 h ∘ g −2 0

f (x) = 3x − 5

g(x) = 3 − 2x

Challenge Problem

The graphs of the equaons and are shown in the grid below.

Which of the following best approximates the value of?

Choose 1 answer:

y = f (x) y = g(x)

  • 8 - 6 - 4 2 4 6 8

2

4

6

8

  • 4
  • 6
  • 8

y

x

y = f (x)

y = g(x)

(f ∘ g)(8)

Check

A −

B −

C 0

D 2