anonymous
  • anonymous
Given that f:x-->2x^2 +8x-10 for the domain x>=k, (iv) find the least value of k for which f is one-one, (v) express f^-1(x) in terms of x in this case.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1332848018302:dw|
anonymous
  • anonymous
I don't understand the question...
anonymous
  • anonymous
look in your class notes and find out where it talks about one-to-one functions. They want you to restrict the domain of the function (just a positive quadratic function) so that it's one-to-one. Find out what makes a function not one-to-one and it will be pretty obvious where k needs to be.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Hmm... Never learned one-one, and it's not in the notes of my class or text book :/
anonymous
  • anonymous
I find it suspicious that a course would ask you a question specifically testing if you know about one-one if they didn't tell you what it is...
anonymous
  • anonymous
your job is to find the vertex if the funciton is one to one it should be either strictly increasing (or decreasing)
anonymous
  • anonymous
It's a past paper question.... Learned the basics of everything, but not in debt.
anonymous
  • anonymous
vertex is \[-\frac{b}{2a}\]
anonymous
  • anonymous
Hmm... So, how do you calculate it?
anonymous
  • anonymous
how do you calculate \[-\frac{b}{2a}\]?
anonymous
  • anonymous
Is it - 8/2(2) ?
anonymous
  • anonymous
\[f:x\to 2x^2 +8x-10\] yes it is
anonymous
  • anonymous
Thank you! I remember seeing this somewhere... but It's not in my textbook... hmm strange.
anonymous
  • anonymous
How about part ii?
anonymous
  • anonymous
put \[x=2y^2+8y-10\] and solve for y using the quadratic formula you will not need to use the \[\pm\] in front of the radical because you have restricted the doman
anonymous
  • anonymous
So, do I just solve like 2y^2 +10y-2y -10 2y(y+5)-2(y+5) (2y-2)(y+5)=x? Or how?
anonymous
  • anonymous
no you need to set it equal to zero
anonymous
  • anonymous
so, y=1 or y=-5?
anonymous
  • anonymous
\[x=2y^2+8y-10\] \[2y^2+8y-x-10=0\] use quadratic formula with \[a=2,b=8,c=-x-10\]
anonymous
  • anonymous
try it if you get stuck ask
anonymous
  • anonymous
Hmmm. Let me try... is it -8 + or - the sqrt of ....?
anonymous
  • anonymous
or is it just +?
anonymous
  • anonymous
ok \[a=2,b=8,c=-x-10\] \[y=\frac{-8+\sqrt{64-4\times 2\times(-x-10)}}{4}\]
anonymous
  • anonymous
OK, so just plus. Thanks
anonymous
  • anonymous
\[y=\frac{-8+\sqrt{64+8x+80}}{4}\] \[y=\frac{-8+\sqrt{144+8x}}{4}\]
anonymous
  • anonymous
can simplify this a little
anonymous
  • anonymous
How?
anonymous
  • anonymous
write radical in simplest radical form \[\sqrt{144+8x}=2\sqrt{36+2x}\] then cancel a 2 top and bottom
anonymous
  • anonymous
OK. Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.