A minimum with no derivative The function f(x)=|x| has an...

A minimum with no derivative The function $f(x)=|x|$ has an absolute minimum value at $x=0$ even though $f$ is not differentiable at $x=0 .$ Is this consistent with Theorem $2 ?$ Give reasons for your answer.

Key Concepts

Fermat's Theorem

Fermat's Theorem states that if a function has a local extremum at an interior point and is differentiable at that point, then the derivative at that point must be zero. This theorem establishes a necessary condition for local extrema in differentiable functions, meaning if the function is smooth at a point and achieves a local maximum or minimum there, its derivative will vanish.

Differentiability

Differentiability refers to the existence of a derivative at a given point, which implies that the function has a well-defined tangent at that point. When a function has features like sharp corners or cusps, it is not differentiable at those points, and thus the derivative-based criteria for identifying extrema do not apply.

Local and Absolute Extrema

Local extrema refer to maximum or minimum values in a small neighborhood around a point, while absolute extrema are the highest or lowest values over the entire domain of the function. The theorems concerning derivatives typically address local extrema under the condition of differentiability, so an absolute extremum occurring at a point where the function is not differentiable does not violate those theorems.

Non-differentiable Critical Points

A point where a function is not differentiable can still be an extremum. The necessary condition provided by Fermat's Theorem only applies when the function is differentiable at that point. Therefore, an extremum occurring at a non-differentiable point is not in conflict with the theorem, as the theorem does not claim to cover such situations.

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