1

I had a question where I had to find range of $p$ (given, it is real) in order of $(x+3p+1)^{1/3}-(x)^{1/3}=1$ to have real solutions.

I was wondering if I can take cube on both sides of inquality and solve further but then it occured to me that squaring of square roots always created problems as we weren't sure if a function was stil less than or greater than the other function.

Please help, I know this question holds equality so we can cube here but what if I had a cube root problem could I still cube it?

Gary
  • 31,845
Shashwat Singh
  • 23
  • 5
  • 2
    For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Aug 28 '22 at 19:22
  • Protip: Use $(x)^{1/3}$ in your post (including the title) to obtain $(x)^{1/3}$. – Brian Tung Aug 28 '22 at 19:30
  • (probably useless) Protip: Use $(x)^1/3$ in your post (including the title) to obtain $(x)^1/3$. – peterwhy Aug 28 '22 at 19:32
  • Also \sqrt[3]x is $\sqrt[3]x$. You can take the third power if $f(x),g(x)>0$ and not switch the signs – Тyma Gaidash Aug 28 '22 at 19:43
  • @TymaGaidash what if we are not sure if f(x),g(x)>0? – Shashwat Singh Aug 28 '22 at 19:47
  • I am assuming that $x^{1/3}$ is the inverse function of $x^3.$ The limit of LHS at $\pm\infty$ is equal $0.$ In order to get the value $1$ we have to make sure that the espression on LHS exceeds $1.$ Then we can use the intermediate value theorem. For example plugging in $x=0$ we can notice that it suffices that $3p+1\ge 1,$ i.e. $p\ge 0.$ – Ryszard Szwarc Aug 28 '22 at 19:52
  • You can, indeed, take the cubes of both sides of an inequality. Or, more generally, apply an increasing function (defined in a suitable interval). I discussed this in a different context here. But it seems to me that your inequality is a bit different because there is that extra term $1$. If this poses no problems in the later steps of your solution, proceed with a clear conscience! – Jyrki Lahtonen Aug 28 '22 at 20:02
  • 1
    You will also get better results for your formulas if (after writing $ to start) you write the whole equation and only put another $ to end it after the entire equation has been written. – David K Aug 28 '22 at 20:26

2 Answers2

3

Assuming that $h(x)=x^{\frac 1 3}$ is a real valued function defined for $x \in \mathbb R$, since $l(x)=x^{3}$ is an increasing function we have

$$f(x)^{\frac 1 3}>g(x)^{\frac 1 3}\iff \left(f(x)^{\frac 1 3}\right)^3>\left(g(x)^{\frac 1 3}\right)^3 \iff f(x)>g(x)$$

indeed also $h(x)=x^{\frac 1 3}$ is an increasing function.

Refer also to the related:

For the square root we have that $h(x)=x^{\frac 1 2}$ is a real valued function defined for $x \ge 0$ therefore since $l(x)=x^{2}$ is an increasing function we have

$$f(x)^{\frac 1 2}>g(x)^{\frac 1 2}\implies \left(f(x)^{\frac 1 3}\right)^2>\left(g(x)^{\frac 1 2}\right)^2 \implies f(x)>g(x)$$

but

$$f(x)>g(x) \not \Rightarrow f(x)^{\frac 1 2}>g(x)^{\frac 1 2}$$

because we can't exclude that $g(x)<0$.

user
  • 154,566
3

Solution 1

It is not worthwhile to raise both sides of the inequality to the third power as it stands. It is better to move one term to the other side and then raise to the third power. $$(x+a)^{1/3}=x^{1/3}+1,\qquad a=3p+1$$ After raising to the third power we get the quadratic equation $$3x^{2/3}+3x^{1/3}+1-a=0$$ The equation is solvable iff $$9-12(1-a)\ge 0$$ i.e. $a\ge {1\over 4}.$ Hence $p\ge -{1\over 4}.$

Solution 2

We may restrict to $a=3p+1>0,$ as the LHS of the inequality is nonpositive for $a\le 0.$ We will apply the formula and the inequality $$u^3-v^3=(u-v)(u^2+uv+v^2),\quad 2(u^2+uv+v^2)\ge u^2+v^2$$ Thus $$u-v\le {2(u^3-v^3)\over u^2+v^2},\qquad u>v$$ Consider the function $$f(x)=(x+a)^{1/3}-x^{1/3}$$ Then ($u=(x+a)^{1/3},\ v=x^{1/3}$) $$0\le f(x)\le {2a\over [(x+a)^{1/3}]^2+[x^{1/3}]^2}$$ Hence $$\lim_{|x|\to \infty} f(x)=0$$ The function $f(x)$ is decreasing on the half-axis $[0,\infty).$ This can be verified by studying the sign of the derivative on $[0,\infty).$ For $x\le 0$ the maximal value is attained at $x=-a/2.$ By the intermediate value theorem the interval $(0,4^{1/3}a^{1/3}]$ is equal to the range of $f.$ Therefore the equation has a solution if and only if $4^{1/3}a^{1/3}\ge 1,$ i.e. $p\ge -{1\over 4}.$

Remark Solution $2$ can be applied to similar question with roots of any odd order, for example $5.$ In that case the solution $1$ is convenient, if applicable at all.

Ryszard Szwarc
  • 30,319