Showing that $x^3-x^2-1$ has only one REAL root.

Question

So, I need to prove that $x^3-x^2-1$ = 0 has only one root.

Here's what I have so far:

We have $f(-2) = -13 < 0, f(2) = 3 > 0$. Because f is a polynomial, it is continuous on all $\mathbb{R}$ and there is c s.t. $-2 < c < 2$ and $f(c) = 0$ by I.V.T. This shows the equation has at least one solution.

Usually from here, people show that $f '(x)$ is strictly increasing, and thus can't have second root. But in this case $f'(x)$ = $3x^2-2x$, and $f'(x) = 0, $ when $x = 0$ and x = $2/3$.

So, I have no idea how to prove it has ONLY one root :( Thanks in advance for your help!

Answer 1

So, you know $f'$ is zero at two points and only two points. We can make use of this information.

Recognize that,

• for $x\in (-\infty,0)$, $f'(x)>0$
• for $x \in (0,2/3)$, $f'(x)<0$
• for $x \in (2/3,\infty)$, $f'(x)>0$

Evaluate $f(x)$ first at each of the zeros of $f'$. Taking a look at the graph of $f$ (below), we immediately conclude that the $f(x) \in (-2,0)$ at those points. More specifically, $f(0) = -1, f(2/3) = -31/27$. The important bit being, they're less than $0$.

Thus, by the monotonicity of $f$ on these intervals we know the behavior of $f$:

• It ascends to $-1$ when $x\in(-\infty,0)$
• It then descends to $-31/27$ when $x \in (0,2/3)$
• Then we ascend for $x>2/3$, cutting through the $x$-axis and giving us our one real root

Since $f$ is not given any other opportunity to ascend above the $x$-axis, we conclude the uniqueness of this root.

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The graph promised:

Answer 2

You know that $f$ is increasing until $x = 0$, then decreasing till $x = \frac{2}{3}$, then increasing again till the rest of time. But $f(0)$ is negative, so the graph can only cross the $x$-axis once.

Answer 3

You have found that $f(x)$ has two critical points using the first derivative rule. Now find which kind of critical points they are, either by the first derivative test or (easier and more common) the second derivative test.

Then find the values of $f(x)$ at those two points. It will then be clear that the function does not have a zero between those two points and also does not have a zero between one of the infinities and one of those points. That means there is only one zero.

Answer 4

$f(-\infty,0)=(-\infty,-1)$ therefore $f(x)<-1, \forall x\in (-\infty,0)$

$f(0,2/3)=(-\frac{31}{27},-1)$ therefore $f(x)<-1, \forall x \in(0,2/3)$

$f(2/3,\infty)=(-31/27,\infty)$ and $f$ strictly increasing therefore vanishes a single time in $(2/3,\infty)$.

Answer 5

Here is a proof that does not use derivatives.

If one consider $f(1)<0$ and $f(2)>0$, one can conclude from the intermediate value theorem that $f$ has a real root $a$ such that $1<a<2$.

Thus for some real numbers $b$ and $c$, one has $$ f(x)=x^3-x^2-1=(x-a)(x^2+bx+c)=x^3+(b-a)x^2+(c-ab)x-ac. $$ So, one has $$ c-ab=0,\quad ac=1,\quad a\neq 0.\tag{1} $$ which implies that $$ a^2b=1.\tag{2} $$ It follows from (1) and (2) that $$ f(x)=(x-a)(x^2+\frac{x}{a^2}+\frac{1}{a}).\tag{3} $$

Now, it is not difficult to see that $$ \frac{1}{a^4}-\frac{4}{a}<0\tag{4} $$ since $a>1$. By (3) and (4), $f$ cannot have any other real root.

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[Note:] (4) is equivalent to $$ 1-4a^3<0\quad \Leftrightarrow\quad 4a^3>1. $$

Answer 6

You only need to restrict the possible domain where a real root can occur.

To do so, note that

• $x^3-x^2-1 = 0 \Leftrightarrow x^3 = x^2+1 \color{blue}{\geq 1}$
• It follows that any real solution can only exist for $\color{blue}{x \geq 1}$

Now, you can reason as usual:

• $f(1) <0, f(2) >0$ and $f$ is strictly increasing for $x\geq 1$.
• It follows that there can be only one real root.

Answer 7

You look for the zero(s) of function $$f(x)=x^3-x^2-1$$ Its first derivative $$f'(x)=3x^2-2x=x(3x-2)$$ cancels for $x=0$ and $x=\frac 23$. By the second derivative test, $x=\frac 23$ corresponds to a minimum; then $x=0$ corresponds to a maximum.

Now $f(0)=-1$ and $f\left(\frac{2}{3}\right)=-\frac{31}{27}$ and, by inspection $f(1)=-1$ and $f(2)=3$. So, only one real root soewhere between $1$ and $2$.