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anonymous
 one year ago
Help with two problems..
anonymous
 one year ago
Help with two problems..

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking graph A

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For this one ive gotten to y= square root x +13 , y</ 13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The square root is only over x

ahvei
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still need help on the second one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0? The second one is not multiple choice. I dont know where you just got C from.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need help on the f(x)= (x13)^2 , x</ 13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont want the answer told to me. Id like to be shown how to solve it.

ahvei
 one year ago
Best ResponseYou've already chosen the best response.0oh is it the tyoe of question you will have to type an answer out for??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes? If you look at the picture it shows that the final answer must be typed in... @phi can you help me with this one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can show my work up to where im stuck at if that will help.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes but my example lost me and i'm stuck at y=  square root x +13 , y</ 13

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im at the part where I found what the restriction and inverse equal.. Restriction = y</ 13 Inverse= y=  square root x +13

phi
 one year ago
Best ResponseYou've already chosen the best response.1in the original y= (x13)^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= (x13)^2 we get 0 when we "put in" 0 in the inverse we get y= 13 another test x= 1 in y=(x13)^2 give y= (113)^2 = 196 we put 196 into the inverse y=  sqr(196)+13 = 14 +13 = 1 we get back 1 so it works

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 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

phi
 one year ago
Best ResponseYou've already chosen the best response.1Do you know the domain and range of the original y= (x13)^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Domain of f= Range of f&1 = ( infinity sign,13]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Following my example it should be [0, infinity sign) but I have to complete part A to be sure.

phi
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1the inverse function switches the domain and range in other words, the domain of f inverse is the range of f

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um okay.. That's part C.. I still need help completing part A.

phi
 one year ago
Best ResponseYou've already chosen the best response.1you mean \[f^{1}(x)= \sqrt{x}+13, \ \ x\ge 0\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I followed the example.. f^1 =  square root x +13 ; x>/0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the graph but I didn't need help with that part.. Just part A had me stuck.

phi
 one year ago
Best ResponseYou've already chosen the best response.1** 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)

phi
 one year ago
Best ResponseYou've already chosen the best response.1or perhaps I don't understand your question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Um I'll type out what I mean ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0y</ 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 <

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why is the sign for the final answer different from the one that it ends with x</0

phi
 one year ago
Best ResponseYou've already chosen the best response.1This 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
 one year ago
Best ResponseYou've already chosen the best response.1If it does not show that, it is a typo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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
 one year ago
Best ResponseYou've already chosen the best response.1Definitely 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
 one year ago
Best ResponseYou've already chosen the best response.1if you try to find the square root of 1 on your calculator is will complain

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so it was an error by Pearson.

phi
 one year ago
Best ResponseYou've already chosen the best response.1a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its 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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