anonymous
  • anonymous
Need Help! Find where the function f(x)=x+sqrt(1-x) is increasing and decreasing.
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1439260678254:dw|
anonymous
  • anonymous
step one find the derivative
anonymous
  • anonymous
|dw:1439260757824:dw|

Looking for something else?

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

More answers

anonymous
  • anonymous
let me know when you get \[f'(x)=1-\frac{1}{2\sqrt{1-x}}\]
anonymous
  • anonymous
I'm just having trouble trying to solve for x
anonymous
  • anonymous
ok lets back up a second
anonymous
  • anonymous
exponential notation is great for computing certain derivatives, but it totally sucks for doing actual numeric computation
anonymous
  • anonymous
the square root function is a very very very common function so you should commit its derivative to memory and not screw around with rational exponents \[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] like memorizing \(6\times 9=54\)
anonymous
  • anonymous
and via the chain rule \[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]
anonymous
  • anonymous
I didnt learn it that way, I was taught to set it up with the 1/2 in the front and -1/2 as the exponent value. But if doing it the way above helps me solve the problem, ill stick with it
anonymous
  • anonymous
math teachers love to put the square root function on a test, quiz etc so while your colleages are converting to exponents, then using the power rule etc, you should go right to \[f'(x)=1-\frac{1}{\sqrt{1-x}}\]
anonymous
  • anonymous
yeah the power rule works for sure, but the derivative never changes
anonymous
  • anonymous
so once you know it, you know it like \(8\times 7=56\) it is always the same and yes, you need to do any computation at all with a derivative, you have to get rid of the rational exponent and write what it actually is
anonymous
  • anonymous
|dw:1439261266064:dw|So would the function look like this
anonymous
  • anonymous
now you can solve \[1-\frac{1}{2\sqrt{1-x}}=0\] to find the critical points etc you cannot do it with \[1-\frac{1}{2}(1-x)^{-\frac{1}{2}}\]that form i useless
anonymous
  • anonymous
yeah what you wrote now we can do it
anonymous
  • anonymous
Do we multiple the 1 by the denominator of the other fraction?
anonymous
  • anonymous
you want to set it equal to zero and solve probably a good first step
anonymous
  • anonymous
oh kk
anonymous
  • anonymous
notice that the deriviative has domain \(x<1\) which is not surprising because the original function has domain \(x\leq 1\)
anonymous
  • anonymous
also notice that the derivative is strictly decreasing
anonymous
  • anonymous
So does x equal 3/4 ?
anonymous
  • anonymous
which means all that is left to do is solve \[1-\frac{1}{2\sqrt{x-1}}=0\]
anonymous
  • anonymous
yes it does
anonymous
  • anonymous
Okay, so then I would make a number line and test points on the left and the right and see where it is postive and negative which should give me where the function is inc or dec?
anonymous
  • anonymous
so \(f'(x)<0\) i.e is negative if \(x\geq \frac{3}{4}\) and \(f'(x)>0\) i.e positive of \(x\leq\frac{3}{4}\)
anonymous
  • anonymous
yeah you can do that if it is not obvious
anonymous
  • anonymous
then translate to increasing decreasing for \(f\) and you are done
anonymous
  • anonymous
quick quiz \[\frac{d}{dx}[\sqrt{1-x^2}]=?\]
anonymous
  • anonymous
so (-infinity, 3/4) is incr and decreasing from (3/4, infinity)
anonymous
  • anonymous
oh no hold the phone
anonymous
  • anonymous
increasing on \((-\infty, \frac{3}{4})\) is right
anonymous
  • anonymous
1/2*sqrt(1-x^2) *-2
anonymous
  • anonymous
1/(2*sqrt(1-x^2))*-2
anonymous
  • anonymous
close the numerator should be \(-2x\)
anonymous
  • anonymous
and of course the two's cancel
anonymous
  • anonymous
Oh yea, was in a rush
anonymous
  • anonymous
ok lets get back to your question
anonymous
  • anonymous
\[f(x)=x+\sqrt{1-x}\]right?
anonymous
  • anonymous
is it undefined from 3/4 to infinity
anonymous
  • anonymous
don't write that is increasing on \((\frac{3}{4},\infty)\) or your teacher will think you are daft
anonymous
  • anonymous
the domain is \[(-\infty, 1]\] right?
anonymous
  • anonymous
right so we just dont include the 3/4 to infinity part, right?!
anonymous
  • anonymous
so it is only increasing on \[(\frac{3}{4},1)\]
anonymous
  • anonymous
yeah you include it , but dont to to infinity
anonymous
  • anonymous
three fourth is less than one, so there is an interval over which it increases
anonymous
  • anonymous
right!
anonymous
  • anonymous
and that's because of the domain that you were explaining before
anonymous
  • anonymous
oops is said "increase" i meant "decrease"
anonymous
  • anonymous
yea I figured!
anonymous
  • anonymous
ok want to see a picture?
anonymous
  • anonymous
sure
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=x%2Bsqrt%281-x%29 click on "real valued plot" otherwise you will get complex plot which is not what you want
anonymous
  • anonymous
cool, thanks
anonymous
  • anonymous
yw \[\frac{d}{dx}[\sqrt{x^2+2x-1}]=?\]
anonymous
  • anonymous
1/(2*sqrt(x^2+2x-1) * 2x+2)
anonymous
  • anonymous
right, better knows as \[\frac{x+1}{\sqrt{x^2+2x-1}}\] easy right?
anonymous
  • anonymous
yea, not bad
anonymous
  • anonymous
so no more screwing around with rational exponents when dealing with square roots!!

Looking for something else?

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