Jtdylktuy

As the question states, how do I prove that this equation has at least one root?
I tried doing the derivative, but it isn't quite conclusive as it just states the direction of the graph (if it's growing or decreasing), and it doesn't prove it crosses the origin (has one root)
How do I prove it?

let $f(x) = x^6 - 5x + 2$

$f(1) = -3$

$f(0) = 2$

There is at least one root between $0$ and $1$

$f(2)= 56$

There must also be at least one root between $1$ and $2$

In general, to show that $f(x)$ has a root (and as an added bonus find a bound for that root), you can try out different values of $x$. If you can find real numbers $a$ and $b$ with $f(a)<0<f(b)$, then you know that there must be some $c$ in $(a,b)$ with $f(c)=0$. This result is known as the intermediate value theorem, a well known fact in real analysis applying to continuous functions $f:[a,b]tomathbb R$ ($a$ and $b$ are real numbers). Here, just test the values $a=0$ and $b=1$ like other answers have suggested to get your proof.

Your best bet is the use of the intermediate value theorem, assuming you want a calculus-based approach.

Show that there exists $x_1,x_2$ such that $f(x_1) < 0$ and $f(x_2) > 0$. Then by the intermediate value theorem, as $f$ is clearly continuous, there exists a $c in (x_1,x_2)$ such that $f(c) = 0$ - and thus, $f$ has a root!

(You can test arbitrary $x_1,x_2$, picking at random if you choose. The intermediate value theorem doesn't tell you how to pick the $x_1,x_2$, just what happens if such points exist.)