Help with two problems..

- Destinyyyy

Help with two problems..

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- schrodinger

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- Destinyyyy

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- Destinyyyy

I was thinking graph A

- DanJS

yes

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- Destinyyyy

Alright thank you.

- Destinyyyy

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- Destinyyyy

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

- Destinyyyy

The square root is only over x

- ahvei

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

- Destinyyyy

Still need help on the second one.

- Destinyyyy

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

- Destinyyyy

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

- Destinyyyy

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

- ahvei

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

- Destinyyyy

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?

- Destinyyyy

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

- Destinyyyy

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

- Destinyyyy

Im at the part where I found what the restriction and inverse equal..
Restriction = y

- phi

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

- Destinyyyy

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

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- Destinyyyy

doing*

- phi

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

- Destinyyyy

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

- phi

and what is the range?

- Destinyyyy

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

- phi

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

- phi

**(-14)^2= 196

- phi

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

- Destinyyyy

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

- Destinyyyy

Yes I know that.

- phi

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

- phi

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

- phi

Here is a graph

##### 1 Attachment

- Destinyyyy

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

- Destinyyyy

I'm confused by one part.. The last step it squares both sides and flipped the sign /0

- Destinyyyy

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

- phi

** The last step it squares both sides and flipped the sign /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)

- phi

or perhaps I don't understand your question?

- Destinyyyy

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

- Destinyyyy

y/ 0
(square root x) ^2 >/ 0^2
x not <

- Destinyyyy

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

- phi

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 >=)

- phi

If it does not show that, it is a typo

- Destinyyyy

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

- phi

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

- phi

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

- Destinyyyy

Okay so it was an error by Pearson.

- Destinyyyy

Alright thanks

- phi

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

- Destinyyyy

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.

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