Find the domain and range of the function ( f(x) = sqrt{x^2 - 9} )

Question

Accepted Answer

Step:1 ：The function ( f(x) = sqrt{x^2 - 9} ) is a square root function, thus the expression under the square root must be greater than or equal to zero.Step:2 ：We set the expression under the square root equal to zero and solve for x: ( x^2 - 9 = 0 ) which gives us ( x = -3 ) or ( x = 3 ). This indicates that the function is defined for ( x leq -3 ) or ( x geq 3 ).Step:3 ：The range of the function is all possible output values. Since we are taking the square root of a number, the output values will always be greater than or equal to zero.

Answer

Step: 1 ：The function ( f(x) = sqrt{x^2 - 9} ) is a square root function, thus the expression under the square root must be greater than or equal to zero.Step: 2 ：We set the expression under the square root equal to zero and solve for x: ( x^2 - 9 = 0 ) which gives us ( x = -3 ) or ( x = 3 ). This indicates that the function is defined for ( x leq -3 ) or ( x geq 3 ).Step: 3 ：The range of the function is all possible output values. Since we are taking the square root of a number, the output values will always be greater than or equal to zero.

Find the domain and range of the function $f (x) = \sqrt{x^{2} - 9}$

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Popular Problems

Find the domain and range of the function f(x)=x2−9

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Popular Problems

Find the domain and range of the function $f (x) = \sqrt{x^{2} - 9}$