If f(x) is a polynomial of degree three with leading coefficient 1 such that $f(1)=1$, $f(2)=4$, $f(3)=9$then prove $f(x)=0$ has a root in interval $(0,1)$.

This is a reframed version of "more than one option correct type" questions. I could identify all the other answers but this one got left out.

My Attempt:
From the information given in the question, the cubic equation is $$x^3-5x^2+11x-6=0$$

Now I don't know how to prove the fact that one of the roots lies in the interval $(0,1)$. As it is a cubic equation I can't even find the roots directly to prove this.

  • 2
    $\begingroup$ If you can show two values $x_1,x_2$ in $(0,1)$ such that $f(x_1)>0$ and $f(x_2)<0$ you are done because $f(x)$ is continuous. $\endgroup$
    – Jesús Ros
    Jan 3 '17 at 9:00
  • 1
    $\begingroup$ A question: with just three given conditions, how did you determine all the four unknowns in $f(x) = ax^3 + bx^2 + cx + d$? $\endgroup$ Jan 3 '17 at 9:01
  • $\begingroup$ @FreezingFire given a=1 (with leading coefficient 1) $\endgroup$
    – oshhh
    Jan 3 '17 at 9:03
  • $\begingroup$ @Osheen I considered that in the three given conditions: $a=1$, $f(1) = 1$ and $f(2)=4$. Did I miss some other condition? $\endgroup$ Jan 3 '17 at 9:04
  • 1
    $\begingroup$ How did you get $f(x) = x^3 - \tfrac{14}{5}x^2+\tfrac{22}{5}x-\tfrac{8}{5}$? Why can't $f(x) = x^3-5x^2+11x-6$ or $f(x) = x^3-6x^2+14x-8$? $\endgroup$
    – JimmyK4542
    Jan 3 '17 at 9:05

You have that: $$f(0) = c,\ f(1)=1,$$ Since you have shown that $c<0$, and since $f$ is continuous, by the intermediate value theorem it must pass through zero at some point in $[0,1]$.


Now that you've added the constraint $f(3) = 9$, the cubic polynomial you found no longer meets the constraints. So you still need to find $f(x)$.

Note that $g(x) = f(x)-x^2$ is also a cubic polynomial with leading coefficient $1$, and $g(1) = g(2) = g(3) = 0$.

Hence, $f(x)-x^2 = g(x) = (x-1)(x-2)(x-3)$, and thus, $f(x) = (x-1)(x-2)(x-3)+x^2 = x^3-5x^2+11x-6$.

Then, since $f(0) = -6 < 0 < 1 = f(1)$, by the IVT, $f(x)$ has a root in $(0,1)$.

  • $\begingroup$ Too many silly mistakes...thank you $\endgroup$
    – oshhh
    Jan 3 '17 at 9:22

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