Show $\;f(x) = x^{20}-70x^3+1\;$ has zero in $\;[0, 1]$

Question

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.)

Answer 1

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

Prompts:

• 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 $[0, 1]$

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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.

Answer 2

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 $f(0) = a_0$,

and

1. $f(1) < 0 $ since $f(1) = \sum_{i=0}^n a_i$.