Question

ihk

Answer 1

Isn't this a quadratic equation problem, so you would use the formula (-b+- sqrt(b^2 - 4ac))/(2a)

Answer 2

the H(t) at least.. the F(x) seems easier

Answer 3

but i don't know what fixed points are

Answer 4

oh, i think I get it.. let me double check with wiki haha

Answer 5

haha ok

Answer 6

this is diffeq right?

Answer 7

diffeq?

Answer 8

never mind. i thought you were finding the fixed points of both functions together.. So, because H(t) is not at all related to f(x), then your fixed points are going to be whatever roots you find.
Ex. H(t) = 3t^2 +18t-6
after solving for t, you get t= -3-sqrt(11) and t= 11-sqrt(3)
so the fixed points would be (0, (-3-sqrt(11))) and (0, (11-sqrt(3)))

Answer 9

H(t) = 3t^2 + 18t -6 Is a parabola that opens at the top, if by fixed points you mean points that lie on that graph their are an infinite number of such points. For example when t=0 then H(t)=-6, when t=1, H(t)=15 and so on. The roots where H(t) =0 are:\[-3-\sqrt{11}\]\[t=\sqrt{11}-3\]

Answer 10

While irrational, they are still real. Fixed.....well they aint gonna move.

Answer 11

fixed points are points whose x-coordinate equals to the y-coordinate.

that is what you get out is whatever you put int.

to find fixed points, set f(x) = x

3t^2+18t−6 = t
solve for t

(5-3x)/4 = x
solve for x

Answer 12

@sourwing Thanks, I think I may be learning something here. Solving the two equations that you provided gives me t=1/3, t=-6
and x=5/7. Is this two problems? Is t=1/3 and H(t)=1/3 and t=-6 and H(t)=-6 for one function and x=5/7 while F(x)=5/7 ?

Answer 13

@radar yes, these are two separate problems

Answer 14

so the fixed points for H(t) are 1/3 and -6 and for F(x) it is 5/7?

Answer 15

yup they are right thanks

Answer 16

Why is a graph a useful source of information?