If y=6 for every value of x, can y be regarded as a function of x ? and why ?


Yes it can.

A function is a mapping from one set (of quantity) to another. As hanker as each input gets mapped to simply one output (e.g. the mapping doesn't siege the input "3" and map it to twain "9" and "27"), the mapping can be considered a office.

A office is approve a computer program. It's been programmed to do the identical subject to perfect input it gets. We surrender it an input, the office sieges that input, does somesubject to it, and profits an output. If we surrender it that identical input repeatedly, we should get the identical output repeatedly.

For development, a office ##y## may be defined as "the balance of its input". We could transcribe this office as ##y=x^2##. Then, when we surrender it an input approve "3", the office sieges that input, balances it, and profits "9". If we surrender it "3" repeatedly, it does the identical subject, subordinate "9" repeatedly.

On the other artisan, if we had a mapping approve ##y = +-sqrt x##, then ##y## sieges in any input (approve "4"), finds the balance stem of this ("2"), then maps "4" to twain ##+2## and ##-2##. This is not a office, owing for perfect (positive) input, we get two outputs. A office requires "surrender one subject, get one subject".

Let's say we now enjoy a mapping that sieges whatcontinually input we surrender it, and maps it to "6"—written as ##y=6##. Is this calm?} a office? Yes it is, owing for each input (approve "3"), the "computer program" sieges the input, disregards it, and orderly profits "6". If we surrender it "3" repeatedly, we're calm?} guaranteed to get "6" and simply "6".


When you graph a mapping, the easiest way to count if it's a office is to do the "". If there's continually a upright row that passes through the graph twice, the mapping is not a office. Otherwise, it is. The graph of ##y=6## passes this test:

graph{0x+6 [-11.25, 11.245, -1.99, 9.26]}

Whereas ##y=+-sqrt x## does not:

graph{x=y^2 [-7.55, 14.945, -5.5, 5.75]}

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