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Show that the given function has exactly one root in given interval. Consider $f(x)= x^4+3x+1$ on the interval $[-2,-1]$.

My try:

By intermediate value theorem there must be at least one real root in the given interval.

Suppose, consider that there are 2 or more real roots in $[-2,-1]$. Hence by Rolle's theorem there must be at least one real $x_0$ such that $f ' = 0$ in $[-2,-1]$ but $ f '=0$ at $x=-\sqrt[3] {\frac {3}{4}}$ , which doesn't belong to the given interval. Hence we have a contradiction. Hence there is exactly one real root in given interval.

I want to know if there are any other standard methods to solve such problems without using contradiction. Because this method is cumbersome when it is very difficult to find the solution of the first derivative to check whether it lies in the given interval or not. Thanks for any help.

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    $\begingroup$ You’re using contradiction here for all of the wrong reasons here. This is literally a simple application to the Intermediate Value Theorem. $\endgroup$ – DaveNine Feb 18 '18 at 15:26
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    $\begingroup$ @DaveNine can you please elaborate how $\endgroup$ – Rohan Shinde Feb 18 '18 at 15:28
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Since $f(-1)=-1<0$ and $f(-2)=11>0$, by the Intermediate Value Theorem we may conclude that there is at least a zero in $(-2,-1)$.

Now notice that $f'(x)=4x^3+3$ and $f''(x)=12x^2\geq 0$ which imply that $f'$ is increasing. Since $f'(-1)=-1$ we have that $f'$ is negative and $f$ is strictly decreasing in $(-\infty,-1]$. Then it follows that the above zero has to be unique.

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  • $\begingroup$ Does this method anyone help when $f(x)=\frac {1}{1-x} + \sqrt {x+1} -3.1$ in the interval $(-1,1)$ $\endgroup$ – Rohan Shinde Feb 19 '18 at 6:33
  • $\begingroup$ @Manthanein Yes, $f$ is the sum of strictly increasing function and therefore $f$ is strictly increasing too. Moreover at $-1$ and at $1^-$ the function $f$ has opposite signs. $\endgroup$ – Robert Z Feb 19 '18 at 6:43
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Without calculus, first note that $\,f(x)= x^4+3x+1\,$ cannot have real positive roots by Descartes' rule of signs. It follows that not all roots can be real, since $\,4\,$ negative roots cannot add up to a sum of $\,0\,$ as given by Vieta's relations.

Then, let $\,z=x+2\,$ so that $\,x \in [-2,-1] \iff z \in [0,1]\,$. Substituting $\,x=z-2\,$ into the equation gives $g(z)=f(z-2)=z^4 - 4 z^3 + 6 z^2 - z - 1\,$.

Again by Descartes' rule of signs $\,g\,$ has exactly $\,1\,$ negative root, and $\,3\,$ or $\,1\,$ positive roots. Since not all roots are real, this excludes the case of $\,3\,$ positive roots, and since $\,f(0) \lt 0 \lt f(1)\,$ the unique positive root in $\,z\,$ must be in $\,(0,1)\,$ i.e. there is a unique root $x = z-2 \in \,(-2,-1)\,$.

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Its derivative $f^\prime(x)=4x^3+3$ vanishes exactly one time and $f(x) \to +\infty$ as $x\to \pm \infty$. Now $f(-1)<0<\min(f(-2),f(0))$, hence by the intermediate theorem and the previous reasoning there is exactly one root in $(-2,-1)$ and exactly one root in $(-1,0)$.

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