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We need to prove that the equation $x^4=3x^3 +1$ has exactly two roots (using the tools Mean Value Theorem and Intermediate Value Theorem ) Now I have done the following work : it is easy to see using the IVT that $f(x)= x^4 -3x^3 -1$ has at least two roots (for example $f(0)=-1<0$, $f(4) > 0$, $f(-4)>0$ ,and so we deduce that there are roots in the intervals $(-4,0)$ and $(4,0)$ ). However here the second derivative is not always positive so I have been unable to apply the MVT to derive the necessary contradiction ...I need to show that is has exactly 2 real solutions...

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  • $\begingroup$ using the tools Mean Value Theorem and Intermediate Value Theorem In this case it would be fairly straightforward to use Descartes' rule of signs instead. $\endgroup$ – dxiv Sep 14 '16 at 5:13
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Hint: by setting $f(x)=x^4-3x^3-1$ we have $f'(x)=x^2(4x-9)$, hence $0$ and $\frac{9}{4}$ are the only stationary points for $f(x)$. $x=0$ is an inflection point and $x=\frac{9}{4}$ is a relative minimum: it follows that $f(x)$ is decreasing on $\left(-\infty,\frac{9}{4}\right)$ and increasing on $\left(\frac{9}{4},+\infty\right)$. Since $\lim_{x\to\pm\infty}f(x)=+\infty$ and $f\left(\frac{9}{4}\right)<0$, $f(x)$ has exactly two real roots.

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  • $\begingroup$ Hi: could you explain the last sentence..... $\endgroup$ – herashefat Sep 14 '16 at 5:04
  • $\begingroup$ @herashefat: if a continuous function $g$ is decreasing (increasing) over the interval $[a,b]$ and $g(a)g(b)<0$, $g$ has exactly one root in $(a,b)$. $\endgroup$ – Jack D'Aurizio Sep 14 '16 at 5:06

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