An assertion is made about a function f that is defined on a...

An assertion is made about a function $f$ that is defined on a closed, bounded interval. If the statement is true, explain why. Otherwise, sketch a function $f$ that shows it is false. (Note: $|f|$ is defined by $|f|(x)=|f(x)| .)$
If $f$ is continuous, then $|f|$ is continuous.

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Key Concepts

Continuity

Continuity is a fundamental property of functions where small changes in the input lead to small changes in the output. Specifically, a function is continuous at a point if, for any arbitrarily small tolerance in the output, there is a corresponding small tolerance in the input such that every input within that tolerance results in outputs that differ by no more than the specified amount. This property is crucial when analyzing behavior on closed, bounded intervals, ensuring no sudden jumps or breaks in the function's behavior.

Absolute Value Function

The absolute value function, defined by sending any real number to its non-negative magnitude, is itself a continuous function. Despite being defined piecewise, the absolute value function does not have any discontinuities. This continuity means that when it is applied to other functions, it can preserve the continuity under composition.

Composition of Continuous Functions

When two functions are continuous, their composition is also continuous. In the context of the problem, since f is continuous and the absolute value function is continuous, the composite function defined by applying the absolute value to f (i.e., |f|(x) = |f(x)|) is also continuous. This principle underpins the assertion in question.

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