Math, asked by Prateek9464, 27 days ago

The range of f(x) = 1/√2-x is ( 0 and infinity )​

Answers

Answered by shimona63
4

Answer:

The given real function is f(x)=x−1

It can be seen that x−1 is defined for (x−1)≥0

i.e., f(x)=(x−1) is defined for x≥1

Therefore the domain of f is the set of all real numbers greater than or equal to 1 i.e.,

the domain of f =[1,∞)

As x≥1

⇒(x−1)≥0

⇒x−1≥0

f(x)≥0

Therefore the range of f is the set of all real numbers greater than or equal to 0

i.e., the range of f =[0,∞)

Answered by LaeeqAhmed
2

 f(x) =  \sf  \frac{1}{ \sqrt{2 - x} }

 \sf \purple{  let}

y =  \frac{1}{ \sqrt{2 - x} }

 \sf \purple{range}

 \implies {y}^{2}  =  \frac{1}{2 - x}

  \implies 2 - x =  \frac{1}{ {y}^{2} }

 \implies  x = 2 -  \frac{1}{ {y}^{2} }

 \frac{1}{ {y}^{2} }  \ne0

 \therefore y ∈ \: \R- \{0 \}

 \orange{ \therefore  \sf range = \R- \{0 \} }

HOPE IT HELPS!

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