4

I've recently started learning calculus. Someone told me that some equations can be solved using derivatives. I was interested about that and I wanted to learn how to do that. As an example problem, I tried to solve an equation using derivatives. Here is the equation:

Problem: Solve for $x \in \mathbb R$: $$(x^3-4)^3=\left(\sqrt[3]{(x^2+4)^2}+4\right)^2.$$

(The problem is taken from this question, which was posted a few days ago. I answered the question with an algebraic solution and now I want to know about the solution which uses calculus.)
Here is my solution to the problem. I want to know whether my solution is correct or not and I also have some questions regarding the solution.

My solution:
It is easy to see that $x\geq \sqrt[3]4$ for equality. And we notice that $x=2$ is a solution.
Let $f(x)=(x^3-4)^3$ and $g(x)=\left(\sqrt[3]{(x^2+4)^2}+4\right)^2$ defined on $[\sqrt[3]4,+\infty)$. Then, $f'(x)=9x^2(x^3-4)^2$ and $$g'(x)=\frac{8x}{3}(\sqrt[3]{x^2+4})+\frac{8x}{3\sqrt[3]{x^2+4}}.$$ Clearly, $f'(x)>g'(x)$ in the interval.
The graph of $f$ is more increasing than $g$ after the equality. So, $f$ and $g$ is can't hold equality for $x>2$, which implies $x=2$ is the only solution to the equation.

The solution says there is no equality for $x>2$. But how to know there is no equality points in $[\sqrt[3]4,2)$? I need an explanation for that. (It would be easy for me if the answer is in terms of graphs of the functions as I don't know much calculus.)
And is this kind of solution to an equation suitable for contests i.e. olympiads (I want to know this because I participate in various contests.)?

Note: The above solution has some errors. I need to prove that $f(x)$ and $g(x)$ can't be equal except $x=2$. My workings are not correct as figured out in comments.

Oshawott
  • 4,028
  • I would say you need to justify more why $f' > g'$. I imagine this is acceptable in contests, but to make sure I would suggest for you to contact the organizers. – user56202 Jul 09 '21 at 15:45
  • Assuming it is true that $f’(x)-g’(x)>0$ so $f-g$ is strictly increasing and thus can have only one zero. – Thomas Andrews Jul 09 '21 at 15:55
  • But your computation for $f’(x)$ is obviously wrong, since $f’(x)$ must be $3x^2\cdot 3(x^3-4)^2$, so must be divisible by $x^2$ and all the terms must be divisible by $9.$ – Thomas Andrews Jul 09 '21 at 15:58
  • Also, $f’(\sqrt[3]4)=0<g’(\sqrt[3]4),$ so the inequality can’t be right on the whole interval. It might be true for $x\geq2,$ which is enough to see there is no solution $>2.$ – Thomas Andrews Jul 09 '21 at 16:01
  • @ThomasAndrews As $f(x)=x^9-12x^2+48x-64$, $f'(x)=9x^8-24x+48$ (sorry if I am missing something). And I assumed $f'(x)>g'(x)$ because it seemed true to me as $f'(x)$ is an eighth degree polynomial and $g'(x)$ involves cube roots. – Oshawott Jul 09 '21 at 16:12
  • $$f(x)=x^9-12x^6+48x^3-64.$$ You forgot that the term is $x^3$ in the later terms. @Unknown But the chain rule was enough for me to figure you were wrong. – Thomas Andrews Jul 09 '21 at 16:14
  • @ThomasAndrews I've just edited this issue. Thanks for pointing that out. – Oshawott Jul 09 '21 at 16:23
  • You haven't clarified which interval $f'>g'.$ It is not all of $(\sqrt[3]4,\infty).$ @Unknown – Thomas Andrews Jul 09 '21 at 16:57
  • @ThomasAndrews This is also not clear to me. How do we prove that $f$ and $g$ can't be equal for $x> 2$? I need help. – Oshawott Jul 09 '21 at 17:19
  • If $f'>g'$ for $x\geq 2,$ this means $h=f-g$ is increasing, so $h(x)>h(2)=0$ for $x>2.$ But that assumes we've shown $f'>g'$ for $x>2.$ So far, we've merely asserted that. @Unknown – Thomas Andrews Jul 09 '21 at 19:01
  • @Unknown Contest problems can be one-trick ponies and I don't think this one was "designed" to be solved this way. $,f-g,$ is not increasing on $,[\sqrt[3]{4}, 2),$, in fact $,f'-g',$ has a $0$ in that interval. And proving it by brute force leads to an algebraic equation of degree $27$, doable by CAS but not by hand. – dxiv Jul 10 '21 at 06:23

1 Answers1

1

Taking instead

$$ x^3 = 4 + ((x^2 + 4)^{\frac 23} + 4)^{\frac 23} = f(x) $$

$f(x)$ has the properties:

$$ \cases{ f(x) = \mathcal{O}\left(|x|^{\frac 89}\right)\\ f'(x) = \frac{x}{g(x)},\ \ \ g(x)\gt 0\\ f(x) \ge 4+\left(4+2 \sqrt[3]{2}\right)^{2/3} } $$

so $x^3$ and $f(x)$ only intersect at one point. As $x=2$ is a solution, this occurs at the intersection point.

Cesareo
  • 36,341