Prove $x^7+3x^5+1$ has exactly one real root using Bolzano's theorem and the MVT.

Question

Prove $f(x)=x^7+3x^5+1$ has exactly one real root using Bolzano's theorem and the MVT.

What I did:

$f(-1)=-3$

$f(0)=1$

As $f$ is continuous, there exists a $c \in (-1,0) /f(c)=0$

Then computed $f'(x)=7x^6+15x^4$.

But $f'\ngtr0 \forall x\in\mathbb{R}$ and $f'\nless0 \forall x\in\mathbb{R}$.

So how can I solve this?

Answer 1

We have $f'(x) = 7x^6 + 15x^4 = x^4(7x^2+15) > 0$ except at $x=0$. This gives us that $x=0$ is a point of inflection. Hence, the function is increasing everywhere except at $x=0$. Hence, there are can be only one root at the most. Further, we have $\lim_{x \to -\infty} f(x) = -\infty$, $\lim_{x \to \infty} f(x) = \infty$ and the function is continuous. Hence, there has to be exactly one root.

Answer 2

Let $x_0$ denote one of the roots that must exist in $(-1,0)$. If $x_1$ is another root, then $f'(\xi)=0$ for some $\xi$ between $x_0$ and $x_1$ (Rolle or MWT). Since $f'(x)=0$ only for $x=0$, we conclude that $x_1>0$. But then again by the MWT, there exists $\eta$ between $0$ and $x_1$ such that $f(x_1)-f(0)=(x_1-0)\cdot f'(\eta)$, i.e., $f'(\eta)=-\frac{1}{x_1}<0$. But $f'(x)\ge 0$ for all $x$.

Answer 3

This is to address your comment where you said you didn't know that you can conclude that $f$ is increasing everywhere because $f' = 0$ at $0$ and greater than $0$ everywhere else.

You know that $f$ is strictly increasing at all points $x < 0$ and strictly increasing at all points $x>0$ because $f' > 0$ everywhere else. Now since $f' \geq 0$ everywhere, $f$ is non-decreasing everywhere. Now consider $\epsilon > 0$. Then $0<\frac{\epsilon}{2}<x$. Then since $f$ is non-decreasing everywhere, $f(-\frac{\epsilon}{2}) \leq f(0) \leq f(\frac{\epsilon}{2})$.

Also, $f(-\epsilon) < f(-\frac{\epsilon}{2})$, since $-\epsilon < -\frac{\epsilon}{2} < 0$ and $f$ is strictly increasing for all $x<0$. Similarly, $f(\frac{\epsilon}{2}) < f(\epsilon)$.

So for any $\epsilon > 0$, $f(-\epsilon) <f(-\frac{\epsilon}{2}) \leq 0 \leq f(\frac{\epsilon}{2}) < f(\epsilon)$. This implies $f(-\epsilon) < 0 < f(\epsilon)$ for any positive $\epsilon$.

Now note that any $x<0$ can be represented as $-\epsilon$ for some $\epsilon >0$ and any $x>0$ can be represented as $\epsilon$ for some $\epsilon > 0$. So for any $x < 0$, $f(x) < f(0)$ and for any $x > 0$, $f(0) < f(x)$.

So now prove that $f$ is strictly increasing, i.e. if you consider any pair of points $x < y \in\mathbb{R}$, then $f(x) < f(b)$. Consider the different cases, e.g. $x <0 $ and $ y < 0$, $x <0$ and $y>0$, $x <0$ and $y=0$, etc.