Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can we use the Intermediate Value Theorem to show that the function $$\;f(x) = x^{20}-70x^3+1\;$$ has a zero in the interval $\;[0, 1]\,$?

(To use the theorem I need to show that the function satisfies all required assumptions.)

share|cite|improve this question
What is your understanding of the intermediate value theorem? What are you struggling with? – muzzlator Apr 4 '13 at 22:49
Man, I just spent ten minutes trying to multiply it by $x^{14}$ to turn it into a quadratic in $x^{17}$ before realizing I forgot about the constant term and it was actually cubic. The intermediate value theorem was a way better idea. That'll teach me to be clever. – Jack M Apr 4 '13 at 23:12

Recall: A "zero" of a polynomial $f(x)$ is a value $x_i$ such that $f(x_i) = 0$.


  • Is $\;f(x) = x^{20} - 70x^3 + 1\;$ continuous on $[0, 1]\;?\;\;$

    • Just recall what you know about the continuity of polynomials on $\mathbb R$.

  • What is $f(0)\;?\;\;$ What is $f(1)\;?\quad$.

  • What does the Intermediate Value Theorem tell you with respect to whether there exists a $c \in [0, 1],\;$ such that $\;0 \lt c \lt 1\;$ and $f(c) = 0$? If you can justify (and it is indeed possible to justify) that there exists such a $c$, then $x_i = c$ is a zero of $f(x)$.

Note: you don't have to actually find the value of the zero; only that such a zero exists in the interval $[0, 1]$

Comment: It would be good for you to include, in your post, what is your understanding of the Intermediate Value Theorem so you can ensure that the conditions for its application are met. You'll want to be able to know the theorem well enough so you can recall it as needed, and state it in your own words.

share|cite|improve this answer
How did I miss this nice answer? +1 – Amzoti May 17 '13 at 1:23

More generally, if $f(x) = \sum_{i=0}^n a_i x^i$, $a_0 > 0$, and $\sum_{i=0}^n a_i < 0$, then $f$ has a root between $0$ and $1$.

To see why,

  1. $f$ is continuous since it is a polynomial,

  2. $f(0) > 0$ since $f(0) = a_0$,


  1. $f(1) < 0 $ since $f(1) = \sum_{i=0}^n a_i$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.