Prove that $f(x)=m$ has three distinct real roots for $m\in(0,8)$

Question

We have $f:\mathbb{R}\rightarrow\mathbb{R},f(x)=x^5-5x+4$ and we need to show that $\forall m\in(0,8)$, $f(x)=m$ has three distinct real roots.

How can I prove it?

Answer 1

Consider $g(x)=f(x)-m$. $g'(x)=5x^{4}-5=5(x^{4}-1)=5(x^{2}+1)(x^{2}-1)$. So g has extrema at $\pm 1$. Notice by the sign of $g'$, $g$ is increasing on $(-\infty,-1]$ and $[1,\infty)$ and decreasing on $[-1,1]$. $g(-1)=8-m$ and $g(1)=-m$. Now if $m \in (0,8)$, $g(-1)>0$ and $g(1)<0$ and you can use the fact that $g(x) \to \infty as x \to \infty$. Added later: we use that $\lim\limits_{x \to -\infty} g(x)=-\infty$. So we can find a unique root strictly smaller than $-1$, and a unique root between $-1$ and $1$, and one root strictly bigger than 1.

Answer 2

Hint: note that $x^5$ is odd and make conclusions on the behavior of the function when $|x|$ is sufficiently large, show that $f(x)$ has two extremas, which satisfy that the maxima is $\geq 8$ and the minima $\leq 0$.

Answer 3

Graphing the function shows that there are three distinct real roots to the equation $f(x) = m$ when $m \in (0,8)$.

Now, to prove this formally, we utilise calculus. We have $f'(x) = 5x^4 - 5 = 5(x^4 - 1)$ and hence the turning points of the function are at $f'(x) = 0 \implies x = \pm 1$ which are the only real solutions to $f'(x) = 0$.

Since $f(-1) = 8$ and $f(1) = 0$ and looking at the signs of $f'(x)$ we can see that $f(x)$ is increasing when $x\leq 1$ and when $x \geq 1$ whilst it is decreasing in the range $[-1, 1]$. Nothing also that $g(x) \to \infty$ as $x \to \infty$ shows that if $m \in (0,8)$ then $f(x) = m$ has three distinct real roots. One in the interval $x \leq 1$, another in the interval $x \geq 1$ and the last in $[-1,1]$. This agrees with our graph.

Answer 4

Solving $f'(x) = 0$ we find that $f$ has a local maximum when $x = -1$, i.e. at $(-1,8-m)$ and a local minimum when $x = 1$, i.e. at $(1,-m)$. Also studying the sign of $f'$ we can conclude that $f$ increases for $x\in (-\infty,-1)\cup (1,+\infty)$ and decreases for $x\in (-1,1)$.

Since $m\in (0,8)$ then the maximum is above $y=0$ and the minimum is below $y=0$.

Using Bolzano's and Rolle's Theorems we can conlude that there's exactly one root in $(-\infty,-1)$, exactly one in $(-1,1)$ and exactly one in $(1,+\infty)$.