Question

determine the inputs that yield the minimum value: F(x)=sqrt[3]{2x^2-8x+22}?????

Answer 1

have you had calculus?

Answer 2

no..i am in precal

Answer 3

\[f(x)=\sqrt{3}*(2x^2-8x+22)\]
\[f'(x)=\sqrt{3}(4x-8)\]
set f' equal to zero
\[\sqrt{3}(4x-8)=0, x=2\]

Answer 4

ok so since you haven't cal
i will do this algebra way also

Answer 5

\[F(x)\sqrt[3]{2x^2-8x+22} \] just wanna make sure we r both seeing the same problem

Answer 6

\[f(x)=\sqrt{3}*2(x^2-4x)+\sqrt{3}*22=2\sqrt{3}(x^2-4x+(\frac{-4}{2})^2)+22\sqrt{3}-2\sqrt{3}*(\frac{-4}{2})^2\]
\[=2\sqrt{3}(x^2-4x+4)+(22-2)\sqrt{3}=2\sqrt{3}(x-2)^2+20\sqrt{3}\]
so we get the min x=2

Answer 7

so you have another probem to
\[f(x)=\sqrt[3]{2x^2-8x+22}\]

Answer 8

so you don't know how to do the derivative right?

Answer 9

i do not

Answer 10

this was the problem from the beginning

Answer 11

so is the min. value h?

Answer 12

\[f(x)=\sqrt{3}(2x^2-8x+22)\]

Answer 13

minimum value will be at the vertex of the quadratic. forget the cube root. the quadratic is a parabola that faces up. will have a minimum value at
\[x=-\frac{b}{2a}\]

Answer 14

x=2 is the min then

Answer 15

you have
\[2x^2-8x+22\]
\[-\frac{b}{2a}=\frac{8}{4}=2\]

Answer 16

no my myininaya 2 is not the minimum value

Answer 17

2 is the input that will get the minimum value yes? minimum means minimum of function, not x that gives it

Answer 18

yessss!! i get as far as getting the 2 but when i plug in i must do something wrong bcuz my answer is the matching up with the correct answer

Answer 19

\[F(2)=\sqrt[2]{2\times 2^2-8\times 2+22}\]
\]

Answer 20

rather the cube root. i get the cube root of
\[8-16+22=-8+22=14\] so minimum should be
\[\sqrt[3]{14}\]

Answer 21

2.41 rounded

Answer 22

oh lordamercy i guess i should read the problem. it says

determine the inputs that yield the minimum value:

Answer 23

it wants the input. and the input is 2!

Answer 24

the book says (2,2.4) so u were still right.

Answer 25

jeez louise

Answer 26

i didn't say the minumum value was 2
i said the minumum was 2

Answer 27

?

Answer 28

the minumum value is 20sqrt{3}

Answer 29

achieves a minimum at (2,2.4)

Answer 30

my master, minimum value is the same as the minimum. output, not input. on the other hand problem asked for input, so i guess it wanted 2 as the answer.

Answer 31

if that is what the book says the book is trash

Answer 32

minimum value means just that. the minimum value. not a coordinate

Answer 33

no
the value refers you the y

Answer 34

exactly

Answer 35

cant argue with u there satel, ive found a hand full of mistakes in this book, so the answer would b 2 afterall?

Answer 36

the min value would be 20sqrt{3}

Answer 37

the answer to the question

"determine the inputs that yield the minimum value?"

is 2

Answer 38

the answer to the question

"what is the minimum value?"

is
\[\sqrt[3]{14}\]

Answer 39

i honeslt dont know what function we are looking at
there has been mention of two different functions
\[f(x)=\sqrt{3}(2x^2-8x+22)\]
and
\[f(x)=\sqrt[3]{2x^2-8x+22}\]

Answer 40

the answer to

"what is the coordinate of the lowest point on the curve of
\[y=F(x)\] ?"

is\[ (2,2.41)\]

Answer 41

i did the first one

Answer 42

@mininaya i have been assuming it is the second one

Answer 43

no myin it was the same function we didnt understand e/o at the beg.

Answer 44

so it was the first one

Answer 45

it is the 2nd

Answer 46

\[F(x)=\sqrt[3]{2x^2-8x+22}\] yes?

Answer 47

ha ha satael73 u were doin the correct one, myi we had a misunders.

Answer 48

the input would have been the same in either case for the same reason. the output would of course be different

Answer 49

either way sate. i believe my professor would like the answer (2,2.4) so i thank u 4 wrkin it out step by step

Answer 50

i wish i knew what problem we were doing lol

Answer 51

ha ha myin. i thank u too! i appreciate it!

Answer 52

you are welcome, so long as you tell your professor that you have it on good authority that "minimum" or "maximum" of a function refers to an output, not a coordinate!

Answer 53

my mininaya please tell me you haven't been here since 8 this morning!

Answer 54

ha ha i will do so, just as soon as my semester is over and my grade is finalized ha ha

Answer 55

i always understood if you don't say the word value you are talking about the x
if you talking about value , than you are talking about y