The local minimum value of the function f'(x) | vedclass

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Question

(C) Given the function $f(x) = 3 + |x|$.We know that the derivative of $|x|$ is given by:$f'(x) = \frac{d}{dx}(3 + |x|) = \frac{d}{dx}(|x|) = 	ext{sg

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given the function $f(x) = 3 + |x|$.We know that the derivative of $|x|$ is given by:$f'(x) = \frac{d}{dx}(3 + |x|) = \frac{d}{dx}(|x|) = 	ext{sgn}(x)$Where $	ext{sgn}(x)$ is the signum function defined as:$f'(x) = \begin{cases} 1, & x > 0 \ -1, & x The function $f'(x)$ is undefined at $x = 0$.Looking at the values of $f'(x)$,it takes only two values: $1$ for $x > 0$ and $-1$ for $x The minimum value of this function $f'(x)$ is $-1$.

The local minimum value of the function $f'(x)$,where $f(x) = 3 + |x|$ for $x \in \mathbb{R}$,is:

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