Question

f(x)=(x^3)-(x^2)+x-2 [0,3] f(c)=4
I already found what f(0) and f(3) are

Answer 1

What is the question?

Answer 2

do you know the rational root test? because I'd think you'd need it

Answer 3

"Verify that the Intermediate Value Theorem applies to the indicated interval and find the value of 'c' guaranteed by the theorem".

Answer 4

& I don't need to use the rational root test..

Answer 5

Do you know what the intermediate value theorem states?

Answer 6

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

Answer 7

Yup, so first of all, is f(c)=4 between f(a) and f(b)?

Answer 8

Yeah

Answer 9

So then the intermediate value theorem guarantees that f(c)=4 has a solution for some c.

Answer 10

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.

Answer 11

Right now i have 4=x^3-x^2+x-2, should I subtract the 4 over before substituting in c?

Answer 12

The equation might have more than one solution, but we are only interested of the solution(s) in the interval [0,3].

Answer 13

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.

Answer 14

On an exam you would want to use c in the equation, so 4=c^3-c^2+c-2

Answer 15

okay now I have 4=c^3-c^2+c-2

Answer 16

Do you know how to solve it?

Answer 17

yeah lol

Answer 18

Then solve it and you will have your answer :)

Answer 19

okay now I have 4=c^3-c^2+c-2

Answer 20

@dape i actually don't know what to do from here lmao

Answer 21

Start by subtracting 4 from both sides.

Answer 22

0= c^3 - c^2 + c - 6

Answer 23

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.

Answer 24

Do you understand this?

Answer 25

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.

Answer 26

I don't understand o_o

Answer 27

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.

Answer 28

where did you get those numbers from?

Answer 29

They are factors of \(-6\), you can multiply them together to get \(-6\).

Answer 30

OH Okay I see where the numbers came from, how did you know to use the factors of 6 though?

Answer 31

@dape

Answer 32

Read carefully what I wrote above, essentially it's a sophisticated form of guess.

Answer 33

okay I got f(1) = -5
f(2) = 0
f(3) = 15
but my book says the answer is f(2) = 4
._.

Answer 34

@dape

Answer 35

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.

Answer 36

OH BECAUSE I WAS SOLVING FOR C

Answer 37

i'm so brain dead