How do you find the domain and range of #f(x)= sqrt(x^2x6)#?
Domain is
Range is
Since under square root the polynomial must be positive, the domain is obtained by solving:
you can obtain the zeroes of the polynomial by solving the associated equation:
so the disequation is solved in the external intervals:
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To find the domain and range of ( f(x) = \sqrt{x^2  x  6} ):

Domain: Find the values of ( x ) for which the expression inside the square root is defined.
 Set the expression inside the square root, ( x^2  x  6 ), greater than or equal to zero.
 Solve the inequality ( x^2  x  6 \geq 0 ) to find the values of ( x ) that satisfy it.
 The domain is the set of all real numbers ( x ) that satisfy the inequality.

Range: Find the possible output values of ( f(x) ).
 Since the square root function outputs only nonnegative values, the range of ( f(x) ) will be all real numbers greater than or equal to zero that satisfy the equation ( y = \sqrt{x^2  x  6} ).
 However, to ensure the square root is nonnegative, the expression inside it must be greater than or equal to zero.

Solve the inequality for the range:
 ( x^2  x  6 \geq 0 ) for ( x ).
 Find the values of ( x ) that satisfy this inequality.
 The range consists of all real numbers ( y ) such that ( y \geq 0 ).
Therefore, the domain is all real numbers except for the values that make the expression inside the square root negative, and the range is all real numbers greater than or equal to zero.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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