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Prove that: If $m<-2$ or $m>2$ ( $m$ is a parametric) then $f(x)=x^{3}-\frac{3}{2}m^{2}x^{2}+32=0$ has exactly three different roots satisfying: $x_{1}<0<x_{2}<x_{3}$

Firstly, $f(x)$ is a continuous function on $\mathbb{R}$

$f(0)=32$, $\lim_{x\to -\infty } f(x)=- \infty$

$\Rightarrow \exists x_{1}<0: f(x_{1})<0$ $\Rightarrow f(0).f(x_{1})<0$

After this step, I can't find another value of $f$ to prove the rest of the problem . Please help me!

dmtri
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Trần Tuấn
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2 Answers2

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Since $f(0) > 0$ and $\lim_{x\to +\infty} = +\infty$ it suffices to find some $x > 0$ with $f(x) < 0$.

The derivative of $f'(x) = 3x^2 - 3m^2 x$ vanishes at $x= m^2$, and indeed $$ f(m^2) = - \frac 12 m^6 + 32 < - \frac 12 2^6 + 32 = 0. $$

Martin R
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HINT: Try to understand where $f$ is increasing and where it is decreasing. Study the derivative of $f$, this helps you.

Hermione
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