Question

Help with two problems..

Answer 1

I was thinking graph A

Answer 2

yes

Answer 3

Alright thank you.

Answer 4

For this one ive gotten to y=- square root x +13 , y</ 13

Answer 5

The square root is only over x

Answer 6

the first one is A let me look at the second one...do you still need help on the second one of are you finished with it

Answer 7

Still need help on the second one.

Answer 8

? The second one is not multiple choice. I dont know where you just got C from.

Answer 9

I need help on the f(x)= (x-13)^2 , x</ 13

Answer 10

I dont want the answer told to me. Id like to be shown how to solve it.

Answer 11

oh is it the tyoe of question you will have to type an answer out for??

Answer 12

Yes? If you look at the picture it shows that the final answer must be typed in... @phi can you help me with this one?

Answer 13

I can show my work up to where im stuck at if that will help.

Answer 14

Yes but my example lost me and i'm stuck at y= - square root x +13 , y</ 13

Answer 15

Im at the part where I found what the restriction and inverse equal..
Restriction = y</ 13
Inverse= y= - square root x +13

Answer 16

in the original
y= (x-13)^2
because we are squaring , y will always be positive. y>=0
in the "inverse equation" we still want y>= 0
but we renamed y to x and vice versa
\[ y = -\sqrt{x} +13, x \ge 0\]
notice we don't want x to be negative (no sqr of negative numbers)
so it makes sense

as a test, if x=13 in y= (x-13)^2 we get 0
when we "put in" 0 in the inverse we get
y= 13

another test
x= -1 in y=(x-13)^2 give y= (-1-13)^2 = 196
we put 196 into the inverse
y= - sqr(196)+13 = -14 +13 = -1
we get back -1
so it works

Answer 17

Sorry but I dont really understand all that. This is my second time going this type of problem but first with a square root. The picture attached is what I have written down

Answer 18

doing*

Answer 19

Do you know the domain and range of the original
y= (x-13)^2
?

Answer 20

Domain of f= Range of f&-1 = (- infinity sign,13]

Answer 21

and what is the range?

Answer 22

Following my example it should be [0, infinity sign) but I have to complete part A to be sure.

Answer 23

the domain is 13 to -infinity
if we put in 13 we get y=0
if we put in 0 we get y= 13^2= 169
if we put in -1, we get (-14)^2 = 216
so it make sense the range is 0 to +infinity

Answer 24

**(-14)^2= 196

Answer 25

the inverse function switches the domain and range
in other words, the domain of f inverse is the range of f

Answer 26

Um okay.. That's part C.. I still need help completing part A.

Answer 27

Yes I know that.

Answer 28

you mean
\[f^{-1}(x)= -\sqrt{x}+13, \ \ x\ge 0\]

Answer 29

the reason we say x>=0 is because we know the domain of f inverse is the range
of f(x) , and that is [0, +infinity)
i.e. x>=0

Answer 30

Here is a graph

Answer 31

I followed the example.. f^-1 = - square root x +13 ; x>/0

Answer 32

I'm confused by one part.. The last step it squares both sides and flipped the sign </ but then the final answer is shown as x>/0

Answer 33

Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.

Answer 34

** The last step it squares both sides and flipped the sign </ but then the final answer is shown as x>/0***

They are unrelated.

as you know , the square root (for example) of 4 is 2 and -2
both work.
when you are solving for the inverse, when you take the square root you have two possible equations
\[ y= \sqrt{x}+13\\ y= -\sqrt{x}+13\]
we have to pick the one that works

the restriction x>=0 comes from looking at the range of f(x) and using that as the domain. (the domain are the allowed x values)

Answer 35

or perhaps I don't understand your question?

Answer 36

Um I'll type out what I mean ...

Answer 37

y</ 13
-square root x +13 </
-square root x </ o
Multiply by -1 both sides

square root x >/ 0
(square root x) ^2 >/ 0^2
x</ 0

The sign changed on the last one but the answer the sign is > not <

Answer 38

Why is the sign for the final answer different from the one that it ends with x</0

Answer 39

This is how you would do it
\[ y\le 13\\ - \sqrt{x}+13\le 13\\ - \sqrt{x} \le 0 \\ \sqrt{x} \ge 0 \\ x \ge 0\]

(notice when you multiply by -1 , you have to swap the <= to >=)

Answer 40

If it does not show that, it is a typo

Answer 41

Yes I noticed that. But my example flipped it back on the last one. Was it an error? It should have stayed as x>/0

Answer 42

Definitely x>=0
As a check, notice if you let x be negative in the inverse function, for example x=-1,
\[ y= -\sqrt{-1} +13 \]
and we can't find the square root of -1, so y is not a real number. That is a good hint we want x>=0

Answer 43

if you try to find the square root of -1 on your calculator is will complain

Answer 44

Okay so it was an error by Pearson.

Answer 45

Alright thanks

Answer 46

a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.

Answer 47

Its an online system. It has a study area with example and practice problems.. Guess ill have to send them an email to fix the typo. Thank you for your help.