Showing that an equation has a root in an interval

Question

Show that the equation $x^4 - 7x^3 + 1 = 0$ has a root in the interval $[0,1]$.

How would I go about working this out in steps?

Answer 1

Let $f(x)=x^4-7x^3+1$. Note that $f(x)$ is a continuous function.

Since $$f(0)=0^4-7\cdot 0^3+1=1\gt 0$$ and $$f(1)=1^4-7\cdot 1^3+1=-5\lt 0,$$

there exists a real number $\alpha$ such that $$f(\alpha)=0\ \ \text{and}\ \ 0\lt\alpha\lt 1$$ by the intermediate value theorem.

Answer 2

As said consider $$f(x)=x^4-7x^3+1$$ and by inspection you notice that $f(0)\times f(1)<0$; so there is at least a root between $0$ and $1$.

Look at the derivative $$f'(x)=4x^3-21x^2=x^2(4x-21)$$ It it positive if $x > \frac {21}4$. So, in the considered interval, there is only one root.