I need to prove there are $3$ real solutions to $x^5 – 4x + 2 = 0$.
I know $f(-2)$ is negative, $f(0)$ is positive, $f(1)$ is negative, $f(2)$ is positive so that by IVT there are at least $3$ roots. Where I get confused is how to apply Rolle's theorem to show the uniqueness of these roots. I know that there must be a number $c$ such that $f'(c)=0$, but what step do I take after that?
Thanks
Best Answer
The function is continuous, so
$f(-2)<0$ and $f(0)>0$ so there is $-2<x_1<0$ that $f(x_1)=0$.
$f(0)>0$ and $f(1)<0$ so there is $0<x_2<1$ that $f(x_2)=0$.
$f(1)<0$ and $f(2)>0$ so there is $1<x_3<2$ that $f(x_3)=0$.
$-2<x_1<0<x_1<1<x_3<2$
so
$x_1<x_2<x_3$, therefore they are different.
Also, $f'(x)=5x^4-4$, has only two roots, so there cannot be more than 3 roots for $f$ because there is a root of $f'$ between each two consecutive roots of $f$.