[Math] Prove that $f(x)=(e^x-1)(x^2+3x-2)+x$ has exactly one positive root, exactly one negative root and one root at $x=0$.

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Prove that the function

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

has exactly one positive root, exactly one negative root and one root at $x=0$.

My work so far:

$f(0)=0$

Thus, $x=0$ is a root.

For the positive root,

$f(\frac{1}{4})=(e^{\frac{1}{4}}-1)((\frac{1}{4})^2+3(\frac{1}{4})-2)+(\frac{1}{4})
=(e^{\frac{1}{4}}-1)(\frac{1}{16}+\frac{3}{4}-2)+\frac{1}{4}
=-\frac{19}{16}e^{\frac{1}{4}}+\frac{23}{16}
<0$
$f(1)=(e^1-1)((1)^2+3(1)-2)+(1)
=(e^1-1)(1+3-2)+1=2e^1-1
>0$

Thus, there is at least one positive root, by the intermediate value theorem. I have also done the same thing for the negative root, as $f(-4)<0$ and $f(-2)>0$.

But how do I show that there is only one positive root and only one negative root?

Best Answer

(To state my solution more rigorously and clearly, I put my comment into an answer.)

First, it is easy to calculate:

$$ f'(x) = e^x (x^2+5x+1) - 2x -2 \\ f''(x) = (x+1)(x+6)e^x-2 \\ f'''(x)=e^x(x^2+9x+13) $$

Second, we will check how many real roots $f''(x)=0$ has:

  • Case $-\infty<x \leq -1$:

    Calculate $f'''(x)=0$, we get the stationary points of $f''(x)$ are at

    $$ x_{1} = \frac{-9 + \sqrt{29}}{2}, ~ x_{2} = \frac{-9 - \sqrt{29}}{2}$$

    Then evaluate: $f''(x_1)\approx -2.55543$, $f''(x_2)\approx -1.99445$. Besides,

    $$ \lim_{x \rightarrow -\infty} f''(x) = -2$$

    and $f''(-1)=-2$. Thus,

    $$\max_{-\infty < x \leq -1} f''(x)=f''(x_2)\approx -1.99445 < 0$$

    That is, in this case, there's no real root for $f''(x)=0$.

  • Case $x>-1$:

    In this case, it is easy to check $f''(x)$ is monotone increasing and is able to go larger than 0. Thus, there's exactly one real root for $f''(x)=0, x>-1$.

Conclusion: $f''(x)=0$ has exactly one real root.

Finally, we will utilize Rolle's theorem:

(Given proper continuity and differentiability.) If $f'(x)=0$ has exactly $n$ real roots, then $f(x)=0$ at most has $n+1$ real roots.

Why? If we can find $n+2$ real roots for $f(x)=0$, say

$$x_1<x_2<\cdots<x_{n+2}$$

according to Rolle's theorem, there will exist $c_1 \in (x_1,x_2)$, $c_2 \in (x_2,x_3)$, ..., $c_{n+1} \in (x_{n+1},x_{n+2})$, such that

$$f'(c_1) = f'(c_2) = \cdots = f'(c_{n+1}) = 0$$

It will contradict "$f'(x)=0$ has exactly $n$ real roots".

Now, we can see because $f''(x)=0$ has exactly one real root, $f'(x)=0$ will at most have two real roots, and then $f(x)=0$ will at most have three real roots. And you have found the three roots for $f(x)=0$, so there're no more roots.


Note: Be careful of the "at most" statements above. It does NOT mean the equation must have such many roots.

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