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...

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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.