How can i prove how many solutions has this equation:$ x^3-3x+b=0$

Question

How can i prove how many solutions has this equation: $x^3-3x+b=0$ This is what I have done, but I got stuck when I try to prove b for $R+$ and $R-$

If $b=0$ so $f'(x) = x^3-3x -> f'(x) = 0 -> x^3-3x = 0 -> x(x^2-3) -> x1 = 3$ and $x2 = +- √3$

If $b = R+$ and $≠ 0 -> f'(x) = 3x^2-3$ and $f''(x) = 6x$ $f(1) = -2+b$ $f(2) = 2+b$ I do not know if $f(1)$ or $f(2)$ are $> 0$ or $< 0$

Can anyone help me to resolve this problem, thanks

Answer 1

Knowing the shape of cubic curves, at most they can have two "hips". We can find the location of these by locating the maxima and minima of the curve:

$$f'(x) = 3x^2-3 = 0 \implies x = \pm 1$$

which gives local max and min for $b=0$ at $\pm 2$, which gives us three cases graphically:

If $|b| < 2$, then the curve will have three real roots.

If $|b| = 2$ then the curve will have two real roots (with one being repeated).

If $|b| > 2$ then the curve only has one real root.

Answer 2

Define the function $f(x)=x^3-3x$. $f(x)$ has relative extrema of $-2$ and $2$. As such, if $|b|\leq2$, then the equation will have three real roots. Otherwise, the equation will have one real root and two complex roots.