L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 1

L32-Wed-16-Nov-2016-Sec-5-1-Composite-HW32-5-2-Inverse-HW33-Q26

Be sure to go over the key to the exam and try and figure out what you did wrong and, more importantly, WHY you did it.

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 2

•

2 1 1 12 3 3 2 3 22 3

x xh x f g x f g x fxx x x

xx

D:g: x ≠ 0

D:f: x ≠ -3

D:h:

3 2 03 2

23

xx

x

Since g is the input for f we could also look at what value of x will make g = -3:

2

23

23

g xx

xx

This is the same result as above. So, domain of f g x is

2| 0,3

x x x

•

1 2 32 2 63 11

3

xj x g f x g f x g xx

x

D:g: x ≠ 0

D:f: x ≠ -3

D k: all reals So, domain of g f x is | 3x x

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 3

2 2 2 2 2

2

3 2 7 3 4 4 7 12 12 3 7

f g x f g x f x a

x a x ax a x ax a

Since f g x crosses y axis at 68, the point (0, 68) satisfies this equation:

2 2

2 2

2

12 12 3 7

68 12 0 12 0 3 775 3

5

f g x x ax a

a aa

a

2

2

12 60 68

12 60 68

f g x x xor

f g x x x

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 4

Sometimes, we can "decompose" a function. That is, we are given a function that is itself a composition and we find two functions that could make up the composition.

That is, find two functions f and g such that f g x h x

There are two operations here: subtracting 5 from x and squaring.

So, let's define

5g x x and 2f x x . We can see that

2

5 5h x x f x f g x f g x

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 5

5.2 One-to-One Functions & Inverse Functions

Defn: A function is a relation where each input corresponds to exactly one output.

For example, these are all functions (they pass the Vertical Line Test).

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 6

Which of the above are one to one functions?

ANS 3f x x and 1f xx

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 7

Note that if we restrict 2, 0f x x x then it is one-to-one.

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 8

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 9

This function maps 2 onto 3, 5 onto 7, and 8 onto -1.

The inverse does this mapping backwards.

That is, it maps 3 onto 2, 7 onto 5, and -1 onto 8.

The inverse can be written as 1 3,2 , 7,5 , 1,8f x .

From this example, we can see that the domain of f x is the range of 1f x AND

the range of f x is the domain of 1f x .

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 10

The inverse, 1

3xf x says to take the input and divide it by 3. This "undoes" multiplication by 3.

Since a function and its inverse "undo" each other, their composition results in the input. That is,

1 1f f x f f x x

For example, if 3f x x and 1

3xf x then

1 3

3 3x xf f x f x and 1 1 33

3xf f x f x x

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 11

2 3 3 3 332 2 32 32

xf g x f xx x xx

xx x

where x ≠ 0 and

•

3 6 2 6 6 32 3 32 2 23 3 22

32 2 33 322 2

x xx x x x xxg f x g x

x xxx x

where x ≠ 2 or 0

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 12

3

3

3

1 1 34

f x x

f

Thus, the point (1,4) satisfies

3 3f x x .

Therefore, the point (4, 1) will satisfy

1 3 3f x x

1 3

1 3

3

4 4 31

f x x

f

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 13

This means that the graph of a function and its inverse are symmetric with respect to the line y = x.

We can show this by showing that the line connecting (1, 4) and 4, 1) is perpendicular to the line y = x (i.e., its slope is -1), and the distance from (1,4) to the line y = x and the distance from (4, 1) to the line y = x are equal. So, if we are given the graph of a function we can graph its inverse by choosing points that are a reflection about the line y = x.

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 14

If we are given the equation of a function we can find its inverse by simply exchanging the x and y values and then solving for y.

Exchange x and y and then solve for y:

3

1 3

3

3

: 1: 1

11

f x y xf x x y

x yx y

D: all reals

R: all reals

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 15

Exchange x and y and then solve for y:

1

4:2

4:2

42

4 2

f x yx

f x xy

yx

yx

Df: x ≠ –2

D 1f x : x ≠ 0

Rf: y ≠ 0

R 1f x : y ≠ –2

L32-Wed-16-Nov-2016-Sec-5-1-Composites-HW32-5-2-Inverses-HW33-Q26 page 16

1

2 3:4

2 3:4

4 2 34 2 32 4 32 4 3

4 32

xf x yxyf x x

yx y yxy x yxy y x

y x xxy

x

Df: x ≠ –4

D 1f x : x ≠ 2

Rf: y ≠ 2

R 1f x : y ≠ –4