Destinyyyy
  • Destinyyyy
Help with two problems..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Destinyyyy
  • Destinyyyy
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Destinyyyy
  • Destinyyyy
I was thinking graph A
DanJS
  • DanJS
yes

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Destinyyyy
  • Destinyyyy
Alright thank you.
Destinyyyy
  • Destinyyyy
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Destinyyyy
  • Destinyyyy
For this one ive gotten to y=- square root x +13 , y
Destinyyyy
  • Destinyyyy
The square root is only over x
ahvei
  • 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
  • Destinyyyy
Still need help on the second one.
Destinyyyy
  • Destinyyyy
? The second one is not multiple choice. I dont know where you just got C from.
Destinyyyy
  • Destinyyyy
I need help on the f(x)= (x-13)^2 , x
Destinyyyy
  • Destinyyyy
I dont want the answer told to me. Id like to be shown how to solve it.
ahvei
  • ahvei
oh is it the tyoe of question you will have to type an answer out for??
Destinyyyy
  • 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
  • Destinyyyy
I can show my work up to where im stuck at if that will help.
Destinyyyy
  • Destinyyyy
Yes but my example lost me and i'm stuck at y= - square root x +13 , y
Destinyyyy
  • Destinyyyy
Im at the part where I found what the restriction and inverse equal.. Restriction = y
phi
  • 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
  • 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
Destinyyyy
  • Destinyyyy
doing*
phi
  • phi
Do you know the domain and range of the original y= (x-13)^2 ?
Destinyyyy
  • Destinyyyy
Domain of f= Range of f&-1 = (- infinity sign,13]
phi
  • phi
and what is the range?
Destinyyyy
  • Destinyyyy
Following my example it should be [0, infinity sign) but I have to complete part A to be sure.
phi
  • 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
  • phi
**(-14)^2= 196
phi
  • phi
the inverse function switches the domain and range in other words, the domain of f inverse is the range of f
Destinyyyy
  • Destinyyyy
Um okay.. That's part C.. I still need help completing part A.
Destinyyyy
  • Destinyyyy
Yes I know that.
phi
  • phi
you mean \[f^{-1}(x)= -\sqrt{x}+13, \ \ x\ge 0\]
phi
  • 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
  • phi
Here is a graph
1 Attachment
Destinyyyy
  • Destinyyyy
I followed the example.. f^-1 = - square root x +13 ; x>/0
Destinyyyy
  • Destinyyyy
I'm confused by one part.. The last step it squares both sides and flipped the sign /0
Destinyyyy
  • Destinyyyy
Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.
phi
  • 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
  • phi
or perhaps I don't understand your question?
Destinyyyy
  • Destinyyyy
Um I'll type out what I mean ...
Destinyyyy
  • Destinyyyy
y/ 0 (square root x) ^2 >/ 0^2 x not <
Destinyyyy
  • Destinyyyy
Why is the sign for the final answer different from the one that it ends with x
phi
  • 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
  • phi
If it does not show that, it is a typo
Destinyyyy
  • 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
  • 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
  • phi
if you try to find the square root of -1 on your calculator is will complain
Destinyyyy
  • Destinyyyy
Okay so it was an error by Pearson.
Destinyyyy
  • Destinyyyy
Alright thanks
phi
  • phi
a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.
Destinyyyy
  • 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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