Question

find the extreme values AND where they occur:
y = sqrt(-x^2 + 2x +3)

Answer 1

You need to find the critical numbers

Answer 2

y'=?

Answer 3

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.

Answer 4

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

Answer 5

well, this is really the upper half of the circle, because your range was \(y\ge0\)

Answer 6

and my bad, the radius squared is 4, but radius just is 2.

Answer 7

so the minimums are
(1±2,0)

ad maximum is up top
so the minimums are
(1,0+2)

Answer 8

thank you so much! sorry i was unable to respond, my computer was acting up. but i understand the problem now. thank you!

Answer 9

was it algebra or calculus?
(curious)

Answer 10

calculus. i solved for the derivative and got this: .5(-x^2 + 2x + 3)^-.5 (-2x+2)

Answer 11

is that correct?

Answer 12

Yes, you are correct!
(Good chain rule :) )

Answer 13

And with that you need critical numbers

Answer 14

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.

Answer 15

from that, without using the algebra you did above, how would i find the min and max?

Answer 16

you aren't given intervals
(let's say we didn't know it is a circle)

Answer 17

i know i have to solve for 0, but i got stuck :(

Answer 18

Yes set f'(x)=0

Answer 19

\(\color{#000000 }{ \displaystyle 0=\frac{-2x-2}{\sqrt{-x^2+2x+3}} }\)

Answer 20

but wait, it is not corrct

Answer 21

you have a 2 in denominator when u differentiate the root

Answer 22

Am I right, or not?

Answer 23

why is a 2 in the denominator when you take the derivative?

Answer 24

\(\color{#000000 }{ \displaystyle \frac{d }{dx} x^{1/2}=(1/2)x^{-1/2} }\)

Answer 25

via the power rule... right

Answer 26

so now i set (1/2)x^-(1/2) equal to 0?

Answer 27

\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x-2}{2\sqrt{-x^2+2x+3}} }\)

Answer 28

ion our case, correct?

Answer 29

oh, sorry +2

Answer 30

\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)

Answer 31

and set that equal to 0 and solve for x?

Answer 32

yup

Answer 33

\(\color{#000000 }{ \displaystyle 0=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)

Answer 34

x=1

Answer 35

and case (2)

Answer 36

where is f(x) defined, but f'(x) NOT defined?

Answer 37

\(\color{#000000 }{ \displaystyle f(x)=\sqrt{-x^2+2x+3} }\)

\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)

Answer 38

(Hint- in derivative unlike in the f(x), the root cannot be =0)

Answer 39

ah, sorry ignore that^

Answer 40

no, i meant ignore what i said about setting it equal to 0 :) i accidentally copied a post i'd made earlier :)

Answer 41

:)

Answer 42

well, the equation wouldn't work if x = 0, because you cannot divide by 0 ... is that what you meant?

Answer 43

no, if the denominator is 0!
(but you are thinking correctly)

Answer 44

if x=0, denom is not 0

Answer 45

there are two solutions for denominator=0

Answer 46

(by exclamation mark neaxt to 0, "0!" I don't mean zero factorial :))

Answer 47

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?

Answer 48

You can simply set what is inside the square root =0, so I don't really understand your question...

Answer 49

\(\color{#000000 }{ \displaystyle f'(x)=\frac{-2x+2}{2\sqrt{-x^2+2x+3}} }\)

-x²+2x+3 =0

Answer 50

ah! i see.

Answer 51

and solve for x?

Answer 52

yes.

Answer 53

one moment while i figure it out please!

Answer 54

sure, take your time../.

Answer 55

-x^2 + 2x = -3? and then could i combine the -x^2 + 2x to make -2x^3?

Answer 56

WOW

Answer 57

or are they different roots and so cannot be combined?

Answer 58

(sorry, algebra is the biggest math concept i struggle with :)

Answer 59

That is very bad!

Answer 60

Algebra is key to "A" in calculus or any other course in math...

Answer 61

they are not "like terms"

Answer 62

hmmm. that must be why I'm stuck with a B+.... :)

Answer 63

You seem to be good at calculus, but you really need to nail the algebra....

Answer 64

yeah :)

Answer 65

You need to know the entire algebra (look common course outline)

Answer 66

the solutions to -x²+2x+3 are x=-1 and x=3

Answer 67

but, later, please look and make sure you know absolutely all of it.

Answer 68

once you had -x^2 + 2x = -3, what did you do to the left side of the equation to get just x=?

Answer 69

\(\color{#000000 }{ \displaystyle -x^2+2x+3=0}\)
\(\color{#000000 }{ \displaystyle x^2-2x=3}\)
\(\color{#000000 }{ \displaystyle x^2-2x+1=4}\)
\(\color{#000000 }{ \displaystyle (x-1)^2=4}\)
\(\color{#000000 }{ \displaystyle x-1=\pm 2}\)
\(\color{#000000 }{ \displaystyle x=\pm 2+
1}\)

Answer 70

completing the square -ma fav.
but precisely,

Answer 71

\(\color{#000000 }{ \displaystyle (x-1)^2=4}\)
when you take the square root of both sides, you get
\(\color{#000000 }{ \displaystyle|x-1|=2}\)
and that is where we get
\(\color{#000000 }{ \displaystyle x-1=\pm 2}\)

Answer 72

this is via the defition of abs value
\(\color{#000000 }{ \displaystyle \sqrt{r^2}=|r|}\)

Answer 73

so are the solutions still x=-1 and x=3? what does the x=±2+1 represent?

Answer 74

\(\pm2\) says that 2 can be neg or pos

Answer 75

how did you get x = -1 and x=3 from that?

Answer 76

so
\(\pm 2+1\)
read as,

either \(+2+1\) OR \(-2+1\)

Answer 77

3 and -1,
respectively

Answer 78

I wonder how you pass the class at all without algebra.
I a saying this because I want you to know algebra.

Answer 79

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 :(

Answer 80

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?

Answer 81

do i need a third point to determine the max and min?

Answer 82

you have THREE critical number

Answer 83

x=1
(when f'(X)=0)

And when f'(x) -- undefined
x=-1, 3

Answer 84

Plug them into the function f(x), and tell me what do you get for each of them

Answer 85

f(3) = ?
f(1) = ?
f(-1) = ?

Answer 86

(give me exact answers please)

Answer 87

when x = -1, y = 2.4 when x = 3, y = 4.2 when x = 1, y = 2.4

Answer 88

that may not be correct.

Answer 89

where x=-1 and x=3 solutions when we set the part in the root =0?

Answer 90

So how come now you get different results for them now?

\(\color{#000000 }{ \displaystyle f(x)=\sqrt{-x^2+2x+3} }\)

\(\color{#000000 }{ \displaystyle f(-1)=\sqrt{-(-1)^2+2(-1)+3} =\sqrt{-1-2+3}=0 }\)
\(\color{#000000 }{ \displaystyle f(3)=\sqrt{-3^2+2(3)+3} =\sqrt{-9+9}=0 }\)

Answer 91

\(\color{#000000 }{ \displaystyle f(1)=\sqrt{-1^2+2(1)+3} =\sqrt{4}=2 }\)

Answer 92

isn't f(1) = swrt(6) ? --> -1^2=1, 2(1)=2, + 3 --> 1+2+3

Answer 93

noo

Answer 94

isn't f(1) = swrt(6) ? --> -1^2=1, 2(1)=2, + 3 --> -1+2+3

Answer 95

:(

Answer 96

-(1)² is still negative 1

Answer 97

\(\color{#000000 }{ \displaystyle f(1)=\sqrt{-1^2+2(1)+3} =\sqrt{-1+2+3}=\sqrt{4}=2 }\)

Answer 98

there is a difference between
(-1)² and -1²