How many roots are there of function $f(x)=x^4-3x^2+x-1$

Question

Problem: How many (real) roots are there of function $f(x)=x^4-3x^2+x-1$

First, I can directly say since degree is $4$ then there can be at most $4$ roots of functions. Moreover, apparently $f(0)=-1$ and $\lim_ {x \to \pm \infty} = \infty$, so because of the IVT there are at least $2$ numbers $c,d \in \mathbb R$ such that $f(c)=f(d)=0$. One from negative and other one from positive side.

Actually, I can conclude the answer by sketching graph but which tool I can use to show that there are exactly $2$ roots for it without using graph.

Answer 1

(Posted as a CW to avoid any appearance of impropriety, since I voted to reopen the question.)

More general techniques for determining the nature of the roots of a quartic have been mentioned in other comments and answers here, as well as under the related question How to determine number of roots and what type for quartic equations?

However, the particular quartic in the question here also allows for a "shortcut" solution.

• The OP covered half the way by noting that two of the roots must be real.

apparently $f(0)=-1$ and $\lim_ {x \to \pm \infty} = \infty$, so because of the IVT there are at least $2$ numbers $c,d \in \mathbb R$ such that $f(c)=f(d)=0$. One from negative and other one from positive side.

• Let $\,x_i \big|_{i=1,2,3,4}\,$ be the four roots, then $\,\frac{1}{x_i}\,$ are the roots of the reciprocal polynomial, or its opposite $\,-x^4 f\left(\frac{1}{x}\right)$ $= x^4 - x^3 + 3 x^2 - 1\,$. By Vieta's relations: $$\,\sum_{1 \le i \le 4} \frac{1}{x_i^2} = \left(\sum_{1 \le i \le 4} \frac{1}{x_i}\right)^2 −2 \cdot \left(\sum_{1 \le i \lt j \le 4} \frac{1}{x_ix_j}\right) = 1^2−2⋅3 \lt 0$$ It follows that the four roots cannot all be real, therefore the remaining two roots must be non-real complex conjugates.

Answer 2

Doesn't need any calculus...

$P(x): f(x)=x^4-3x^2+x-1 = (x^2-2)^2+(x+0.5)^2-5.25.$

$Q(x): f(x)=x^2(x^2-3)+x-1.$

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$\text{Case 1. } x \geq \sqrt{2}.$

$P(x) \Rightarrow f(x): \text{increasing function, } f(\sqrt{2})=\sqrt{2}-3<0, \displaystyle \lim_{n \to +\infty}f(n) = \infty > 0.$

$\Rightarrow \text{There exists 1 real root for $f(x)=0$ in $[\sqrt{2}, \infty]$.}$

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$\text{Case 2. } x \leq -\sqrt{2}.$

$P(x) \Rightarrow f(x): \text{decreasing function, } f(-\sqrt{2}) = -\sqrt{2}-3<0, \displaystyle \lim_{n \to -\infty} f(n)=\infty>0.$

$\Rightarrow \text{There exists 1 real root for $f(x)=0$ in $[-\infty, -\sqrt{2}].$}$

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$\text{Case 3. } -\sqrt{2} < x < 1.$

$0 \leq x^2 < 2.$

$\Rightarrow -2.25 \leq x^2(x^2-3) \leq 0.$

$-\sqrt{2}<x<1.$

$\Rightarrow -\sqrt{2}-1<x-1<0.$

$\therefore Q(x) \Rightarrow f(x)<0, \text{No real root.}$

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$\text{Case 4. } 1 \leq x < \sqrt{2}.$

$1 \leq x^2 < 2.$

$\Rightarrow 0<(x^2-2)^2 \leq 1.$

$1.5 \leq x + 0.5 < \sqrt{2}+0.5 < 1.5 + 0.5 = 2.$

$\Rightarrow 2.25 \leq (x+0.5)^2 < 4.$

$\therefore P(x) \Rightarrow f(x)<-0.25, \text{No root.}$

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$\therefore f(x)=0 \text{ has only 2 real roots, each in } [\sqrt{2}, \infty] \text{ and } [-\infty, -\sqrt{2}].$

Answer 3

As already suggested in the comments, using the graph, that is maximum, minimum, derivative, etc., with IVT is of course a good canonical way to discuss the number of solutions.

In this case we can also proceed for example as follows, let consider

$$g(x)=f(x)+1=x^4-3x^2+x=x(x^3-3x+1)$$

which has exactly four roots, one trivial at $x=0$ and the three others in the intervals $(-\infty,-1)$, $(0,1)$, $(1,\infty)$. Indeed $h(x)=x^3-3x+1\implies h'(x)=3x^2-3=0\;\text{as}\; x=\pm 1$.

Now we are interested in the extreme value of $g(x)$ between the roots at $x=0$ and at $x\in (0,1)$, which is a local maximum, indeed consider the derivative

$$g'(x)=4x^3-6x+1 \implies g'\left(\frac16\right)>0,\;\;g'\left(\frac15\right)<0$$

with maximum value bounded by

$$g(x)=f(x)+1=x^4-3x^2+x\le \left(\frac15\right)^4-3\left(\frac16\right)^3+\frac15<1$$

therefore $f(x)=g(x)-1$ has exactly $2$ roots in the intervals $(-\infty,-1)$ and $(1,\infty)$.