[Math] Critical value of a piecewise function

calculusderivatives

Consider the following function $f(x)$.

$$f(x)=
\left\{ \begin{array}{ll}
x^2 & , x≤1 \\
|x-2| & , x>1
\end{array}\right.$$

to find the critical value I did the following steps:

  1. Redefine the function without absolute value

$$f(x)=
\left\{ \begin{array}{ll}
x^2 & , x≤1 \\
x-2 & , x>2 \\
-x+2 & , 2>x>1
\end{array}\right.$$

  1. Take the derivative of $f(x)$
    $$f'(x)=
    \left\{ \begin{array}{ll}
    2x & , x≤1 \\
    1 & , x>2 \\
    -1 & , 2>x>1
    \end{array}\right.$$

  2. Find where $f'(x)=0$ or is $undefined$

$f'(x)=0$ when $x=0$, therefore​ $x=0$ is a critical value.

$f'(2)=undefined$ , therefore $x=2$ is also a critical value.

$f'(1)=undefined$ , since $\lim\limits_{x\to 1^-} $ doesn't equal to $\lim\limits_{x\to 1^+} $ , $x=1$ is a critical value.

is this reasoning mathematically correct for proving? If there's anything incorrect let me know,and if you have a better approach for proving please provide your answers​, specially for proving $2$ and $1$ as critical values.

Best Answer

Critical Points are Stationary points and non differentiable points. Your reasoning is correct !

The only think we need to lool in Theese cases that the point we have found out to be critical lies in the domain ( the 1st case). The rest of your 2 cases are correct.

And I should aur one more thing. In case of inflexion point - " They are the point where the 2nd derivative changes sign and need necessarily be 0 "

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