Help with two problems..

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Help with two problems..

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I was thinking graph A
yes

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Alright thank you.
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For this one ive gotten to y=- square root x +13 , y
The square root is only over x
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
Still need help on the second one.
? The second one is not multiple choice. I dont know where you just got C from.
I need help on the f(x)= (x-13)^2 , x
I dont want the answer told to me. Id like to be shown how to solve it.
oh is it the tyoe of question you will have to type an answer out for??
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?
I can show my work up to where im stuck at if that will help.
Yes but my example lost me and i'm stuck at y= - square root x +13 , y
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
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
doing*
  • phi
Do you know the domain and range of the original y= (x-13)^2 ?
Domain of f= Range of f&-1 = (- infinity sign,13]
  • phi
and what is the range?
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
Um okay.. That's part C.. I still need help completing part A.
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
I followed the example.. f^-1 = - square root x +13 ; x>/0
I'm confused by one part.. The last step it squares both sides and flipped the sign /0
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?
Um I'll type out what I mean ...
y/ 0 (square root x) ^2 >/ 0^2 x not <
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
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
Okay so it was an error by Pearson.
Alright thanks
  • phi
a typo. People who look for errors in math books are better at spelling than noticing if an equation makes sense.
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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