How do I prove that a function has real roots?

Question

I want to prove the existence of real roots of a function, not solve the function for the roots. I am aware of discriminant, but that is restricted to quadratic functions. I am aware of the intermediate value theorem, but it can only prove the existence of one and not multiple real roots...

In particular, I am looking to prove that $x^4 - 1102x^3 - 2018 = 0$ has at least $2$ real roots.

Edit: Using the graph, I know the roots are at 1102 and -1.23333. Is using IVT to prove them, using 2 different domains, [1101, 1103] and [-2, -1] a valid proof?

Answer 1

Note that

$$f(x)=x^4-1002x^3-2018\implies f(0)=-2018$$

and $$\lim_{x\to \pm \infty}f(x) = +\infty$$

then refer to IVT.

It is not a proof but a graph is always useful to visualize what is going on

Answer 2

To prove existence of roots of a continuous function, you can exhibit changes of sign.

For instance,

$$f(-10000)>0, f(0)<0, f(10000)>0$$ proves at least two roots.

(Using $\pm\infty$ like gimusi did is also possible.)

Answer 3

Consider the function $$f(x)=x^4 - 1102x^3 - 2018$$ Its derivative $$f'(x)=4x^3-3306x^2=4x^2\left(x-\frac {1653} 2\right)$$ cancels twice, once at $x=0$ and once at $x=\frac {1653} 2$. Notice that $$f\left(\frac{1653}{2}\right)=-\frac{2488686346715}{16}$$ and the second derivative test $$f''(x)=12 x^2-6612 x\implies f''\left(\frac{1503}{2}\right)=2732409$$ shows that this is a minimum.

Since moreover, for large negative $x$, the function behaves as $x^4$, then $\cdots$.