Question

If the functions f(x)=ax^3-bx+1 and g(x)=x^2-2ax-b for x=1 the vertical length beetween the functions is 3 and Cf and Cg intersect with the line ε=2x+1 find a,b

Answer 1

Plugging in x = 1 into both f and g we get:
\[f(1) = a-b+1, g(1) = 1-2a-b\]
now it doesnt say which function is greater, so im going to assume f(x) > g(x) and hope for the best. Since the vertical length between them is 3 at x = 1, we subtract them and set it equal to three:
\[f(1)-g(1) = 3 \Rightarrow (a-b+1)-(1-2a-b) = 3 \Rightarrow 3a = 3 \Rightarrow a = 1\]

Answer 2

f(1)=a-b+1
g(1)=1-2a-b
\[d=\sqrt{(f(1)-g(1))^2}=3\]
\[f(1)-g(1)=3\]
\[(a-b+1)-(1-2a-b)=3\]

Answer 3

lol gj joe

Answer 4

lol, i was on the phone and had a bajillion mistakes in it >.<

Answer 5

looks good to me. what the monkey is Cf and Cg?

Answer 6

im not sure

Answer 7

a random constant times the function was my guess, but i havent started working it out yet

Answer 8

ima do it on paper before i type junk >.>

Answer 9

can't wait to see this...

Answer 10

so i guess we set Cf=2x+1
and set Cg=2x+1

Answer 11

i wonder if we should assume both constants are the same
i think not

Answer 12

Cf and Cg for the graph of f(x) and g(x)

Answer 13

c(x^3-bx+1)=2x+1
and we
have
c'(x^2-2b-b)=2x+1

Answer 14

Thank you very much!! this is a tough one for me!

Answer 15

hey jason what does C mean
is that a constant times the function?

Answer 16

yeah, after playing with it a little, i dont think its a constant anymore lol

Answer 17

i thing it might be craft, i am not sure,that is how we symbolize the graph of a function in greece!!!

Answer 18

\[x^3-bx-2x=0\]
\[x^3-x(b+2)=0\]
\[x(x^2-(b+2))=0\]
\[x=0, x^2=b+2\]
\[x=0, x= \pm \sqrt{b+2}\]
\[x^2-2x-3b-1=0\]
\[x=0=> -3b-1=0\]
\[b=\frac{1}{-3}\]

Answer 19

we have two equations above since both f and g intersect the line e=2x+1

Answer 20

so i did f-(2x+1)=0
and g-(2x+1)=0

Answer 21

are you making this up?

Answer 22

ah, very nice

Answer 23

dont we all "make it up"? :P gotta go with the flow

Answer 24

solved f-(2x+1)=0 this for x
and then plugged x into g-(2x+1)=0 to solve for b (and we already knew a=1 from joe)

Answer 25

also i plug the pretty x into g-(2x+1)=0
not the scary x

Answer 26

Geniusly done!

Answer 27

its definitely correct, i tested those values of a and b to see if at x =1 they were 3 units apart, its good.

Answer 28

\[\color{blue}{\text{myininaya is a hero}}\]

Answer 29

let me know if you have any questions jason?

Answer 30

no understood everything,but just to let you know this exersise is just one of the simple ones (let's say) for our final exams in next years exams in may,so this is just the a start,all this to a 17 years old kids,Anyway, THANK YOU VERY MUCH FOR YOUR HELP

Answer 31

this looks like advanced algebra
lol the college algebra i seen in schools don't ask questions like this

Answer 32

oh you are 17 lol

Answer 33

i wish i was challenge more in high school
i hate my high school

Answer 34

WELL here in greece the last highschool year is really tough!we will be solving exersises with function aprox. 3 books, and make over 30 of practice exams

Answer 35

hated and hate*

Answer 36

seems intense o.O its worth it though

Answer 37

yeah I also wish my high school was more challenging. would have prepared me a lot more for college and beyond.