Polynomial question: find $a$ such that all roots are real.

Question

Original question: let $f:\mathbb{R}\to\mathbb{R},f(x)=\frac {x^2+ax+5}{\sqrt{x^{2}+1}}.$ Find $a$ such that f has 3 distinct local extreme points.

So I differentiated the function and got:$$f'(x)=\frac {x^3-3x+a}{(x^2+1)\sqrt{x^2+1}}$$

So then question is reduced to: find $a$ such that $x^3-3x+a=0$ has all $3$ roots real and distinct.

Answer 1

To find the proper range for $a$, you may look at it as follows:

• $h(x) = x^3-3x+a = x(x+\sqrt{3})(x-\sqrt{3}) + a$
• So, how far can you shift the graph of $c(x) = x(x+\sqrt{3})(x-\sqrt{3})$ upwards and downwards,resp., without the local extrema of $c(x)+a$ reaching the $x$-axis
• local extrema of $c(x)$ are at $x_M = -1$ and $x_m = 1$
• $c(x_M) = 2$ and $c(x_m) = -2 \Rightarrow -2 < a < 2$

Answer 2

Since $x^3-3x+a$ is a cubic polynomial, all you need is for its local extrema to have opposite signs. The extrema occur at the zeros of the derivative, $3x^2-3$, i.e., at $x=\pm1$, with the local maximum at $x=-1$ and the local minimum at $x=1$. So we need

$$(-1)^3-3(-1)+a\gt0\gt1^3-3\cdot1+a$$

or $2+a\gt0\gt-2+a$, which can be rewritten as

$$2\gt a\gt-2$$