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The range of the function f(x)=5/3-x^2

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Solution: For f(x) to be defined , 3-x^2 eq 0

\ herefore Domain of f=R-pmsqrt3

Now, Let y=5/3-x^2Rightarrow x^2=3y-5/y

\ y(3y-5)/y^2. For x to be real x^2geq 0Rightarrow y(3y-5)geq 0\ \ herefore y< 0;or;ygeq 5/3

Hence , range of f=(-infty ,0)cup [5/3,infty )

 

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Deependra Verma

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