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What is the question?
do you know the rational root test? because I'd think you'd need it
"Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of 'c' guaranteed by the theorem".
& I don't need to use the rational root test..
Do you know what the intermediate value theorem states?
If f is continuous on the closed interval between [a,b] and k is any number between f(a) and f(b), then there is at least one number c in [a,b] such that f(c)=k
Yup, so first of all, is f(c)=4 between f(a) and f(b)?
So then the intermediate value theorem guarantees that f(c)=4 has a solution for some c.
So now, if you put in c for x in the function and put it equal to 4 (f(c)=4) and solve for c, you will have your answer.
Right now i have 4=x^3-x^2+x-2, should I subtract the 4 over before substituting in c?
The equation might have more than one solution, but we are only interested of the solution(s) in the interval [0,3].
You should substitute c before, if you want to be thorough, although it doesn't really matter as long as you keep track of your variable names.
On an exam you would want to use c in the equation, so 4=c^3-c^2+c-2
okay now I have 4=c^3-c^2+c-2
Do you know how to solve it?
Then solve it and you will have your answer :)
okay now I have 4=c^3-c^2+c-2
@dape i actually don't know what to do from here lmao
Start by subtracting 4 from both sides.
0= c^3 - c^2 + c - 6
Yes, now since the coefficient in front of the highest order term (\(c^3\)) is 1, any rational solutions must be a divisor of the constant term \(-6\). So you can guess the solution by putting in different factors of \(-6\), one is bound to be a solution. Also the intermediate value theorem guarantees that the solution lies in [0,3], so check which factor of \(-6\) between 0 and 3 is a solution.
Do you understand this?
In other words, when you have an equation like that with whole number coefficients and a 1 in front of the highest order term you can always guess solutions from the factors of the constant term.
I don't understand o_o
When you see a polynomial equation (like yours) with only whole numbers as coefficients it's a good idea to guess what the solution is. The right answer must be a factor of the constant term, in your case the constant term is \(-6\), the factors of \(-6\) is \[\pm1,\pm2,\pm3,\pm6\] You could put in each of these 8 numbers to find which one solves the equation. But you also know from the intermediate value theorem that a solution must exist between 0 and 3 (the interval in the problem), so you only have to consider \[+1,+2,+3\] Now one of these three must be a solution, try each and you will find the answer.
where did you get those numbers from?
They are factors of \(-6\), you can multiply them together to get \(-6\).
OH Okay I see where the numbers came from, how did you know to use the factors of 6 though?
Read carefully what I wrote above, essentially it's a sophisticated form of guess.
okay I got f(1) = -5 f(2) = 0 f(3) = 15 but my book says the answer is f(2) = 4 ._.
If you put 2 into \(c^3 - c^2 + c - 6\) you get 0, so it solves our equation. This means that \(f(2)=4\) as you can check. Try putting 2 into the function and you will see that the book is correct.
OH BECAUSE I WAS SOLVING FOR C
i'm so brain dead