Do all the peaks and valleys of f have f'=0

[AMan24]

I learned in the earlier chapters that peaks and valleys of a fxn have points where f'=0 (i marked them with red x). A few chapters later it said if a fxn has 2 roots, then f'=0 (still the 1st graph).

So does that mean if the graph of a fxn is like the 2nd graph, the peaks and valleys are not f'=0? I drew where i would assume f'=0 with blue circles

 

[Niladri Dan]

If there is a knotch then f'(x) is not equal to zero...

 

[SlowThinker]

All peaks and all valleys have f'(x)=0 but there can be a point where f'(x)=0 but it's not peak or valley. For example f(x)=x^3 at x=0.

The correct wording is that if a function has a double root, not two roots, then f'(x)=0 at that point. Happens for example with f(x)=x^2 at x=0.

 

[Niladri Dan]

All peaks and all valleys have f'(x)=0 but there can be a point where f'(x)=0 but it's not peak or valley. For example f(x)=x^3 at x=0.

The correct wording is that if a function has a double root, not two roots, then f'(x)=0 at that point. Happens for example with f(x)=x^2 at x=0.

Take the case y=-|x| at x=0

 

[SlowThinker]

Take the case y=-|x| at x=0

Yes the rules only apply if the function has a continuous derivative.
With non-continuous functions or derivatives, pretty much anything can happen.

 

[JonnyG]

View attachment 196209

A few chapters later it said if a fxn has 2 roots, then f'=0 (still the 1st graph).

What it should say, or maybe what you meant to say, was that if a function has two roots, say $x_1$ and $x_2$, then there is a point $c$ between $x_1$ and $x_2$ such that $f'(c) = 0$