Question

let f(x)=x^3-x^2+x. show that there is a numbr c such that f(c)=10

Answer 1

I think you would do

10=x^3-x^2+x

Answer 2

Thanks. But wat would you do then

Answer 3

solve for x

Answer 4

i don't believe it is true, since you have a cubic function which goes from \(-\infty\) to \(\infty\)

Answer 5

oooh i see doh
it says show that there IS a number
no problem

Answer 6

any cubic function has range all real numbers, so of course it is true

Answer 7

how do you know its true? could you like, solve the problem for me so I could see and learn frm how you did it?

Answer 8

in your case as \(x\to -\infty\) you know \(f(x)=x^3-x^2+x\to -\infty\) and as \(x\to \infty\) you have \( f(x)=x^3-x^2+x\to \infty\) and since all polynomials are continuous, it must be equal o ten somewhere

Answer 9

another way you could to it is pick some number for \(a\) so that \(f(a)<10\) and some number \(b\) so that \(f(b)>10\) and then say
"since \(f\) is a polynomial it is continuous, so it must take on all values between \(f(a)\) and \(f(b)\)

Answer 10

the point is a cubic function is onto
so it takes on every value