Why do the solutions to the derivatives of polynomials have the same x value as their peaks and valleys? [closed]

I was graphing polynomials and their derivatives, and I noticed that the local maximums and minimums of of polynomials have the same x value as it's derivative's solutions. Is this just a coincidence?

For example, the polynomial, $3x^{5}+x^{4}+0.4x^{3}+x^{2}+2$, has a peak at, $-0.555$. The derivative of that polynomial is, $15x^4+4x^3+1.2x^2+2x$. It has the solution, $x = -0.555$

Note: The question has been changed. The following answers the new question: "Why do local extremas of a differentiable function $f(x)$ occur precisely where $f'(x)=0$?"

Suppose we have a differentiable function $f(x)$, and further suppose that $f(x)$ has a local max (or local min) at $x=c$. Then it must be that the slope of the tangent line to $f(x)$ at $x=c$ is $0$; that is, $f(x)$ has a horizontal tangent at $x=c$.

The derivative $f'(x)$ tells us the slope of the tangent line to $f(x)$ at any $x$-value we want. Therefore, it must be that $f'(c)=0$, since the we know that $f(x)$ has a tangent slope of $0$ at $x=c$.

In summary, if $f(x)$ has a local extrema at $x=c$, then $f'(c)=0$.

(However, if $f'(c)=0$, this does not necessarily mean that $f(x)$ has a local extrema at $x=c$; for example, $f(x)=x^3$ at $x=0$)