Use synthetic division to determine whether the number k is an upper or lower bound (as specified for the real zeros of the function f). k = 2; f(x) = 2x^3 + 3x^2 - 4x + 4; Lower bound?
synthetic is a process, what about the process are you having an issue with?
Yes
show me your work, or possibly ask a question about the process like how to start it, what to use, where did i go wrong ... etc
I actually just got it. My coefficients got mixed up. They are all positive so it's in the upper bound. I'm sorry. But do you think you could help me solve a different question. I have no idea where to start with it.
i might .. let see the question :)
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. f(x) = the quantity x minus seven divided by the quantity x plus three. and g(x) = quantity negative three x minus seven divided by quantity x minus one.
do you recall what the notaiton f(4) means?
also, if you can post a more mathical construction that would be helpful. words and math just dont mix
The questions looks like f(x) = (x-7)/(x-3) and g(x) = (-3x - 7)/(x - 1)
f(g(x)) is just when the x parts of f, are equal to g(x) f(g(x)) = (g(x)-7)/(g(x)-3) (g(x)-7-3+3)/(g(x)-3) 1 - 4/(g(x)-3) we have g(x) defined so plug it in and simplify to see if it reduces to x
or work it into g(x) 1 - 4/[g(x)-3] = x 1 - x =4/[g(x)-3] g(x) - 3 =4/(1-x) g(x) = 4/(1-x) + 3 is this the definition of g(x) ?
Yes?
if you are unsure, then just work the rest of the algebra ...
I have no idea what I'm doing :(
which is why i asked to start with if you knew what the notation f(4) means, if you kow what this notation means then the rest is just process.
I don't know what that is
then you are not prepared to work this problem. spose we have a function f, defined in terms of x: then f(x) is the name of the function if we want to know what the value of f is when x=4, then we can express that by the notation f(x=4), or shorthanded to f(4) now, tell me what this means to you
Nothing. It means absolutely nothing. I'm wasting your time. I'm just going to skip the questions. Thanks anyway.
... wish i could have helped out better then, but good luck all the same :)
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