Need Help! Find where the function f(x)=x+sqrt(1-x) is increasing and decreasing.

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Need Help! Find where the function f(x)=x+sqrt(1-x) is increasing and decreasing.

Mathematics
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|dw:1439260678254:dw|
step one find the derivative
|dw:1439260757824:dw|

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let me know when you get \[f'(x)=1-\frac{1}{2\sqrt{1-x}}\]
I'm just having trouble trying to solve for x
ok lets back up a second
exponential notation is great for computing certain derivatives, but it totally sucks for doing actual numeric computation
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 \(6\times 9=54\)
and via the chain rule \[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]
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
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 \[f'(x)=1-\frac{1}{\sqrt{1-x}}\]
yeah the power rule works for sure, but the derivative never changes
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
|dw:1439261266064:dw|So would the function look like this
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
yeah what you wrote now we can do it
Do we multiple the 1 by the denominator of the other fraction?
you want to set it equal to zero and solve probably a good first step
oh kk
notice that the deriviative has domain \(x<1\) which is not surprising because the original function has domain \(x\leq 1\)
also notice that the derivative is strictly decreasing
So does x equal 3/4 ?
which means all that is left to do is solve \[1-\frac{1}{2\sqrt{x-1}}=0\]
yes it does
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?
so \(f'(x)<0\) i.e is negative if \(x\geq \frac{3}{4}\) and \(f'(x)>0\) i.e positive of \(x\leq\frac{3}{4}\)
yeah you can do that if it is not obvious
then translate to increasing decreasing for \(f\) and you are done
quick quiz \[\frac{d}{dx}[\sqrt{1-x^2}]=?\]
so (-infinity, 3/4) is incr and decreasing from (3/4, infinity)
oh no hold the phone
increasing on \((-\infty, \frac{3}{4})\) is right
1/2*sqrt(1-x^2) *-2
1/(2*sqrt(1-x^2))*-2
close the numerator should be \(-2x\)
and of course the two's cancel
Oh yea, was in a rush
ok lets get back to your question
\[f(x)=x+\sqrt{1-x}\]right?
is it undefined from 3/4 to infinity
don't write that is increasing on \((\frac{3}{4},\infty)\) or your teacher will think you are daft
the domain is \[(-\infty, 1]\] right?
right so we just dont include the 3/4 to infinity part, right?!
so it is only increasing on \[(\frac{3}{4},1)\]
yeah you include it , but dont to to infinity
three fourth is less than one, so there is an interval over which it increases
right!
and that's because of the domain that you were explaining before
oops is said "increase" i meant "decrease"
yea I figured!
ok want to see a picture?
sure
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
cool, thanks
yw \[\frac{d}{dx}[\sqrt{x^2+2x-1}]=?\]
1/(2*sqrt(x^2+2x-1) * 2x+2)
right, better knows as \[\frac{x+1}{\sqrt{x^2+2x-1}}\] easy right?
yea, not bad
so no more screwing around with rational exponents when dealing with square roots!!

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