0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Can you input the function in the question? $\endgroup$ – Archis Welankar Sep 29 '18 at 11:11
  • $\begingroup$ If function is differentiable sometimes nth order derivative of function comes in handy. $\endgroup$ – Archis Welankar Sep 29 '18 at 11:12
  • $\begingroup$ Graphing the function and observing its roots can give you some direction. $\endgroup$ – user512116 Sep 29 '18 at 11:14
  • $\begingroup$ What do you mean by input? Do you mean if it can be input into a calculator? Or if you want a specific function? If it is the latter I am trying to prove that x^4 - 1102x^3 -2018 has 2 real roots. Also, could I ask you to clarify the nth order derivative? $\endgroup$ – Vaishnavi MadhuBhuvaneshwaran Sep 29 '18 at 11:15
  • 2
    $\begingroup$ To add to lulu's comment: The derivative $4x^3-3306x^2$ is coprime to $x^4-1102x^3-2018$, hence we need not deal with roots of higher multiplicity. Therefore, only exactly two or exactly four real roots are possible. But four separate roots would mean that the derivative has three separate roots - whereas the derivative has a double root at $0$ and hence only two seperate roots. Conclusion: Your polynomial has exactly two real roots, both simple $\endgroup$ – Hagen von Eitzen Sep 29 '18 at 11:32
2
$\begingroup$

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 $\cdots$.

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ Thanks for the quote! $\endgroup$ – gimusi Sep 29 '18 at 12:29
1
$\begingroup$

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

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.