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How do you find the domain and range for #y = sqrt(x^2 + 2x + 3)#?

Answer 1

Kennedy Adams

Take note of that

#x^2+2x+3 = x^2+2x+1+2#

#=(x+1)^2+2 >= 2 AA x in RR#

So #sqrt(x^2+2x+3)# is defined #AA x in RR# and the domain is the whole of #RR#.

Also we see that the range is #{y in RR: y >= 2}#. The minimum value #y=2# occurs when #(x+1) = 0#, i.e. when #x=-1#.

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Answer 2

Eva Abramson

To find the domain and range of ( y = \sqrt{x^2 + 2x + 3} ):

Domain: Since the expression under the square root must be non-negative, we need to find the values of ( x ) that make ( x^2 + 2x + 3 \geq 0 ).
- To determine the domain, solve the quadratic inequality ( x^2 + 2x + 3 \geq 0 ).
Range: Since the square root function returns non-negative values, the range of ( y ) will be all non-negative real numbers.
- Hence, the range of ( y ) is ( [0, +\infty) ).

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Answer from HIX Tutor

When evaluating a one-sided 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.