Amenah8
  • Amenah8
find the extreme values AND where they occur: y = sqrt(-x^2 + 2x +3)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
You need to find the critical numbers
SolomonZelman
  • SolomonZelman
y'=?
SolomonZelman
  • SolomonZelman
3 cases when critical numbers occur (1) The x-values that are a solution to 0=f'(x) are critical numbers. (2) The x-values at which f'(x) is undefined, provided f(x) is defined for them. (3) Any closed boundaries are automatically critical numbers.

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SolomonZelman
  • SolomonZelman
If want, algebra \(\color{#000000 }{ \displaystyle y=\sqrt{-x^2+2x+3} }\) \(\color{#000000 }{ \displaystyle y=\sqrt{-x^2+2x-1+4} }\) \(\color{#000000 }{ \displaystyle y=\sqrt{-(x^2-2x+1)+4} }\) \(\color{#000000 }{ \displaystyle y=\sqrt{-(x-1)^2+4} }\) \(\color{#000000 }{ \displaystyle y^2=-(x-1)^2+4}\) circle centered at (1,0) with radius of 4
SolomonZelman
  • SolomonZelman
well, this is really the upper half of the circle, because your range was \(y\ge0\)
SolomonZelman
  • SolomonZelman
and my bad, the radius squared is 4, but radius just is 2.
SolomonZelman
  • SolomonZelman
so the minimums are (1±2,0) ad maximum is up top so the minimums are (1,0+2)
Amenah8
  • Amenah8
thank you so much! sorry i was unable to respond, my computer was acting up. but i understand the problem now. thank you!
SolomonZelman
  • SolomonZelman
was it algebra or calculus? (curious)
Amenah8
  • Amenah8
calculus. i solved for the derivative and got this: .5(-x^2 + 2x + 3)^-.5 (-2x+2)
Amenah8
  • Amenah8
is that correct?
SolomonZelman
  • SolomonZelman
Yes, you are correct! (Good chain rule :) )
SolomonZelman
  • SolomonZelman
And with that you need critical numbers
SolomonZelman
  • SolomonZelman
3 cases when critical numbers occur (1) The x-values that are a solution to 0=f'(x) are critical numbers. (2) The x-values at which f'(x) is undefined, provided f(x) is defined for them. (3) Any closed boundaries are automatically critical numbers.
Amenah8
  • Amenah8
from that, without using the algebra you did above, how would i find the min and max?
SolomonZelman
  • SolomonZelman
you aren't given intervals (let's say we didn't know it is a circle)
Amenah8
  • Amenah8
i know i have to solve for 0, but i got stuck :(
SolomonZelman
  • SolomonZelman
Yes set f'(x)=0
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle 0=\frac{-2x-2}{\sqrt{-x^2+2x+3}} }\)
SolomonZelman
  • SolomonZelman
but wait, it is not corrct
SolomonZelman
  • SolomonZelman
you have a 2 in denominator when u differentiate the root
SolomonZelman
  • SolomonZelman
Am I right, or not?
Amenah8
  • Amenah8
why is a 2 in the denominator when you take the derivative?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle \frac{d }{dx} x^{1/2}=(1/2)x^{-1/2} }\)
SolomonZelman
  • SolomonZelman
via the power rule... right
Amenah8
  • Amenah8
so now i set (1/2)x^-(1/2) equal to 0?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x-2}{2\sqrt{-x^2+2x+3}} }\)
SolomonZelman
  • SolomonZelman
ion our case, correct?
SolomonZelman
  • SolomonZelman
oh, sorry +2
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)
Amenah8
  • Amenah8
and set that equal to 0 and solve for x?
SolomonZelman
  • SolomonZelman
yup
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle 0=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)
SolomonZelman
  • SolomonZelman
x=1
SolomonZelman
  • SolomonZelman
and case (2)
SolomonZelman
  • SolomonZelman
where is f(x) defined, but f'(x) NOT defined?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f(x)=\sqrt{-x^2+2x+3} }\) \(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)
SolomonZelman
  • SolomonZelman
(Hint- in derivative unlike in the f(x), the root cannot be =0)
Amenah8
  • Amenah8
ah, sorry ignore that^
Amenah8
  • Amenah8
no, i meant ignore what i said about setting it equal to 0 :) i accidentally copied a post i'd made earlier :)
SolomonZelman
  • SolomonZelman
:)
Amenah8
  • Amenah8
well, the equation wouldn't work if x = 0, because you cannot divide by 0 ... is that what you meant?
SolomonZelman
  • SolomonZelman
no, if the denominator is 0! (but you are thinking correctly)
SolomonZelman
  • SolomonZelman
if x=0, denom is not 0
SolomonZelman
  • SolomonZelman
there are two solutions for denominator=0
SolomonZelman
  • SolomonZelman
(by exclamation mark neaxt to 0, "0!" I don't mean zero factorial :))
Amenah8
  • Amenah8
so the denominator = 0 when x = .... well, 2 time the sqrt(0) = 0 how can i set what is in the sqrt to equal 0?
SolomonZelman
  • SolomonZelman
You can simply set what is inside the square root =0, so I don't really understand your question...
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\) -x²+2x+3 =0
Amenah8
  • Amenah8
ah! i see.
Amenah8
  • Amenah8
and solve for x?
SolomonZelman
  • SolomonZelman
yes.
Amenah8
  • Amenah8
one moment while i figure it out please!
SolomonZelman
  • SolomonZelman
sure, take your time../.
Amenah8
  • Amenah8
-x^2 + 2x = -3? and then could i combine the -x^2 + 2x to make -2x^3?
SolomonZelman
  • SolomonZelman
WOW
Amenah8
  • Amenah8
or are they different roots and so cannot be combined?
Amenah8
  • Amenah8
(sorry, algebra is the biggest math concept i struggle with :)
SolomonZelman
  • SolomonZelman
That is very bad!
SolomonZelman
  • SolomonZelman
Algebra is key to "A" in calculus or any other course in math...
SolomonZelman
  • SolomonZelman
they are not "like terms"
Amenah8
  • Amenah8
hmmm. that must be why I'm stuck with a B+.... :)
SolomonZelman
  • SolomonZelman
You seem to be good at calculus, but you really need to nail the algebra....
Amenah8
  • Amenah8
yeah :)
SolomonZelman
  • SolomonZelman
You need to know the entire algebra (look common course outline)
SolomonZelman
  • SolomonZelman
the solutions to -x²+2x+3 are x=-1 and x=3
SolomonZelman
  • SolomonZelman
but, later, please look and make sure you know absolutely all of it.
Amenah8
  • Amenah8
once you had -x^2 + 2x = -3, what did you do to the left side of the equation to get just x=?
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle -x^2+2x+3=0}\) \(\color{#000000 }{ \displaystyle x^2-2x=3}\) \(\color{#000000 }{ \displaystyle x^2-2x+1=4}\) \(\color{#000000 }{ \displaystyle (x-1)^2=4}\) \(\color{#000000 }{ \displaystyle x-1=\pm 2}\) \(\color{#000000 }{ \displaystyle x=\pm 2+ 1}\)
SolomonZelman
  • SolomonZelman
completing the square -ma fav. but precisely,
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle (x-1)^2=4}\) when you take the square root of both sides, you get \(\color{#000000 }{ \displaystyle|x-1|=2}\) and that is where we get \(\color{#000000 }{ \displaystyle x-1=\pm 2}\)
SolomonZelman
  • SolomonZelman
this is via the defition of abs value \(\color{#000000 }{ \displaystyle \sqrt{r^2}=|r|}\)
Amenah8
  • Amenah8
so are the solutions still x=-1 and x=3? what does the x=±2+1 represent?
SolomonZelman
  • SolomonZelman
\(\pm2\) says that 2 can be neg or pos
Amenah8
  • Amenah8
how did you get x = -1 and x=3 from that?
SolomonZelman
  • SolomonZelman
so \(\pm 2+1\) read as, either \(+2+1\) OR \(-2+1\)
SolomonZelman
  • SolomonZelman
3 and -1, respectively
SolomonZelman
  • SolomonZelman
I wonder how you pass the class at all without algebra. I a saying this because I want you to know algebra.
Amenah8
  • Amenah8
yeah... :) recently, most of our tests have been finding the derivative, so I haven't had to worry about simplifying too much. last year in precal, i would come in for extra help a lot, but i've always managed to only get B's in math :(
Amenah8
  • Amenah8
so once i have my two x values, i plugged them into the equation to get my extremes (2.4 and 4.2) is that right?
Amenah8
  • Amenah8
do i need a third point to determine the max and min?
SolomonZelman
  • SolomonZelman
you have THREE critical number
SolomonZelman
  • SolomonZelman
x=1 (when f'(X)=0) And when f'(x) -- undefined x=-1, 3
SolomonZelman
  • SolomonZelman
Plug them into the function f(x), and tell me what do you get for each of them
SolomonZelman
  • SolomonZelman
f(3) = ? f(1) = ? f(-1) = ?
SolomonZelman
  • SolomonZelman
(give me exact answers please)
Amenah8
  • Amenah8
when x = -1, y = 2.4 when x = 3, y = 4.2 when x = 1, y = 2.4
SolomonZelman
  • SolomonZelman
that may not be correct.
SolomonZelman
  • SolomonZelman
where x=-1 and x=3 solutions when we set the part in the root =0?
SolomonZelman
  • SolomonZelman
So how come now you get different results for them now? \(\color{#000000 }{ \displaystyle f(x)=\sqrt{-x^2+2x+3} }\) \(\color{#000000 }{ \displaystyle f(-1)=\sqrt{-(-1)^2+2(-1)+3} =\sqrt{-1-2+3}=0 }\) \(\color{#000000 }{ \displaystyle f(3)=\sqrt{-3^2+2(3)+3} =\sqrt{-9+9}=0 }\)
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f(1)=\sqrt{-1^2+2(1)+3} =\sqrt{4}=2 }\)
Amenah8
  • Amenah8
isn't f(1) = swrt(6) ? --> -1^2=1, 2(1)=2, + 3 --> 1+2+3
SolomonZelman
  • SolomonZelman
noo
SolomonZelman
  • SolomonZelman
isn't f(1) = swrt(6) ? --> -1^2=1, 2(1)=2, + 3 --> -1+2+3
Amenah8
  • Amenah8
:(
SolomonZelman
  • SolomonZelman
-(1)² is still negative 1
SolomonZelman
  • SolomonZelman
\(\color{#000000 }{ \displaystyle f(1)=\sqrt{-1^2+2(1)+3} =\sqrt{-1+2+3}=\sqrt{4}=2 }\)
SolomonZelman
  • SolomonZelman
there is a difference between (-1)² and -1²
Amenah8
  • Amenah8
so f(3) = 0?
SolomonZelman
  • SolomonZelman
yes, as I showed...
SolomonZelman
  • SolomonZelman
you can rely on these results for now, but later...
SolomonZelman
  • SolomonZelman
As soon as you can, look at the algebra, and try to nail every concept so that even in dream or in amnesia you would know.... very important system alert ...xD but, seriously, please do...
Amenah8
  • Amenah8
haha, okay :)
SolomonZelman
  • SolomonZelman
f(3)=f(-1)=0 that means they are minimums
SolomonZelman
  • SolomonZelman
f(1)=2 it is maximum
Amenah8
  • Amenah8
so if two values = 0, which would be considered that min? is there only a max (2) ?
SolomonZelman
  • SolomonZelman
Not, zero, but if they yeild the smallest output
SolomonZelman
  • SolomonZelman
not (necessarily), 0, but if they yeild the smallest output
Amenah8
  • Amenah8
but the smallest value i have is 0, and two x's have that (-1 and 3) so which is the min? neither?
SolomonZelman
  • SolomonZelman
I am saying that in this case =0 means minimum, because zero is smallest output.
Amenah8
  • Amenah8
so both are the min?
SolomonZelman
  • SolomonZelman
But I am also saying that the definition of minimum is NOT f( something) =0 but, RATHER f( something) = smallest output
SolomonZelman
  • SolomonZelman
yes, f(3) and f(-1) are the minimums
SolomonZelman
  • SolomonZelman
a minimum of 0, that occurs at x=-1, 3
Amenah8
  • Amenah8
ah! thank you so much for working this out with me. and for staying up so late :)
SolomonZelman
  • SolomonZelman
no problem. good luck with math!
Amenah8
  • Amenah8
thank you!

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