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

- anonymous

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

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- anonymous

|dw:1439260678254:dw|

- anonymous

step one
find the derivative

- anonymous

|dw:1439260757824:dw|

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## More answers

- anonymous

let me know when you get
\[f'(x)=1-\frac{1}{2\sqrt{1-x}}\]

- anonymous

I'm just having trouble trying to solve for x

- anonymous

ok lets back up a second

- anonymous

exponential notation is great for computing certain derivatives, but it totally sucks for doing actual numeric computation

- 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

and via the chain rule
\[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]

- 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

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

yeah the power rule works for sure, but the derivative never changes

- 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

|dw:1439261266064:dw|So would the function look like this

- 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

yeah what you wrote
now we can do it

- anonymous

Do we multiple the 1 by the denominator of the other fraction?

- anonymous

you want to set it equal to zero and solve probably a good first step

- anonymous

oh kk

- anonymous

notice that the deriviative has domain \(x<1\) which is not surprising because the original function has domain \(x\leq 1\)

- anonymous

also notice that the derivative is strictly decreasing

- anonymous

So does x equal 3/4 ?

- anonymous

which means all that is left to do is solve
\[1-\frac{1}{2\sqrt{x-1}}=0\]

- anonymous

yes it does

- 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

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

yeah you can do that if it is not obvious

- anonymous

then translate to increasing decreasing for \(f\) and you are done

- anonymous

quick quiz
\[\frac{d}{dx}[\sqrt{1-x^2}]=?\]

- anonymous

so (-infinity, 3/4) is incr and decreasing from (3/4, infinity)

- anonymous

oh no hold the phone

- anonymous

increasing on \((-\infty, \frac{3}{4})\) is right

- anonymous

1/2*sqrt(1-x^2) *-2

- anonymous

1/(2*sqrt(1-x^2))*-2

- anonymous

close
the numerator should be \(-2x\)

- anonymous

and of course the two's cancel

- anonymous

Oh yea, was in a rush

- anonymous

ok lets get back to your question

- anonymous

\[f(x)=x+\sqrt{1-x}\]right?

- anonymous

is it undefined from 3/4 to infinity

- anonymous

don't write that is increasing on \((\frac{3}{4},\infty)\) or your teacher will think you are daft

- anonymous

the domain is \[(-\infty, 1]\] right?

- anonymous

right so we just dont include the 3/4 to infinity part, right?!

- anonymous

so it is only increasing on \[(\frac{3}{4},1)\]

- anonymous

yeah you include it , but dont to to infinity

- anonymous

three fourth is less than one, so there is an interval over which it increases

- anonymous

right!

- anonymous

and that's because of the domain that you were explaining before

- anonymous

oops is said "increase" i meant "decrease"

- anonymous

yea I figured!

- anonymous

ok
want to see a picture?

- anonymous

sure

- 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

cool, thanks

- anonymous

yw
\[\frac{d}{dx}[\sqrt{x^2+2x-1}]=?\]

- anonymous

1/(2*sqrt(x^2+2x-1) * 2x+2)

- anonymous

right, better knows as
\[\frac{x+1}{\sqrt{x^2+2x-1}}\] easy right?

- anonymous

yea, not bad

- anonymous

so no more screwing around with rational exponents when dealing with square roots!!

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